GIFT  OF 
Dr.   Horace   Ivie 


EDUCATION  DEPT 


TREATISE 


ON 


A  L   G  E   B   R  1. 


BY  ELIAS  LOOMIS,  LLR, 

1 1 

PUOFES80B  OP  NATUBAL  PHILOSOPHY  AND  ASTKONOMY   IN  YALE  COLLEGE,  AND  ATTTHOB 
OF  A  "COUBSK  OF  MATHEMATICS." 


REVISED    EDITION. 


NEW  YORK: 

HARPER   &    BROTHERS,   PUBLISHERS, 
327   TO   335   PEARL    STREET, 

FBANKLIN     SQUARE. 

1886. 


} 


* 


GIFT  OF 


LOGMIS'3  SERIES  OF  TEXT-BOOKS. 


ELEMENTARY  ARITHMETIC.  166  pp.,  28  cents. 
TREATISE  ON  ARITHMETIC.  352  pp.,  88  cents. 
ELEMENTS  OF  ALGEBRA.  Revised  Edition.  281  pp.,  90  cents. 

Key  to  Elements  of  Algebra,  for  Use  of  Teachers.     128  pp.,  90  cents. 
TREATISE  ON  ALGEBRA.    Revised  Edition.    384  pp.,  $1  00. 

Key  to  Treatise  on  Algebra,  for  Use  of  Teachers.  219  pp.,  $1  00. 
ALGEBRAIC  PROBLEMS  AND  EXAMPLES.  258  pp.,  90  cants. 
ELEMENTS  OF  GEOMETRY.  Revised  Edition.  388  pp.,  $1  00. 

ELEMENTS  OF  TRIGONOMETRY,   SURVEYING,  AND  NAVIGATION.    194  pp.,  f  1  00. 
TABLES  OF  LOGARITHMS.     150  pp.,  $1  00. 

The  Trigonometry  and   Tables,  bound  in  one  volume.    360  pp.,  $1  50. 
ELEMENTS  OF  ANALYTICAL  GEOMETRY.    Revised  Edition.    261  pp.,  $1  00. 
DIFFERENTIAL  AND  INTEGRAL  CALCULUS.    Revised  Edition.    309  pp.,  $1  00. 

The  Analytical  Geometry  and  Calculus,  bound  in  one  volume.    570  pp.,  $.1  75. 
ELEMENTS  OF  NATURAL  PHILOSOPHY.    351  pp.,  $1  06. 
ELEMENTS  OF  ASTRONOMY.    254  pp.,  $1  00. 
PRACTICAL  ASTRONOMY.    499  pp.,  $1  50. 
TREATISE  ON  ASTRONOMY.    351  pp.,  $1  50. 
TREATISE  ON  METEOROLOGY.    308  pp.,  $1  50. 


EDUCATION 


,  according  to  Act  of  Congress,  in  the  year  one  thousand  eight  hundred  and 
sixty-eight  by 

HARPER    &    BROTHERS, 
In  the;  Clerk's  Office  of  the  District  Court  of  the  Southern  Didtrict  of  New  York. 


PREFACE. 


THE  stereotype  plates  of  my  Treatise  on  Algebra  having  be 
come  so  much  worn  in  the  printing  of  more  than  60,000  copies 
that  it  had  become  necessary  to  cast  them  aside,  I  decided  to 
improve  the  opportunity  to  make  a  thorough  revision  of  the 
work.  I  therefore  solicited  criticisms  from  several  college  pro 
fessors  who  had  had  much  experience  in  the  use  of  this  book, 
and  in  reply  have  received  numerous  suggestions.  The  book 
has  been  almost  entirely  rewritten,  nearly  every  page  of  it 
Laving  been  given  to  the  printer  in  manuscript.  The  general 
plan  of  the  original  work  has  not  been  materially  altered,  but 
the  changes  of  arrangement  and  of  execution  are  numerous.  In 
the  former  editions,  in  place  of  abstruse  demonstrations,  I  some 
times  employed  numerical  illustrations,  or  deductions  from  par 
ticular  examples.  In  the  present  edition  such  methods  have 
been  discarded,  'and  I  have  aimed  to  demonstrate  with  concise 
ness  and  elegance  every  principle  which  is  propounded. 

This  book  therefore  aims  to  exhibit  in  logical  order  all  those 
principles  of  Algebra  which  are  most  important  as  a  prepara 
tion  for  the  subsequent  branches  of  a  college  course  of  mathe 
matics.  I  have  retained,  with  but  slight  alteration,  a  feature 
which  was  made  prominent  in  the  former  editions,  that  of  stating 
each  problem  twice:  first  as  a  restricted  numerical  problem, 
and  then  in  a  more  general  form,  aiming  thereby  to  lead  the 
student  to  cultivate  the  faculty  of  generalization.  At  the  same 
time  I  have  very  much  increased  the  number  of  examples  in 
corporated  with  each  chapter  of  the  book,  and  at  the  close  have 
given  a  large  collection  of  examples,  to  which  the  teacher  may 
resort  whenever  occasion  may  require. 

The  proofs  of  the  work  have  all  been  examined  by  Prof. 
H.  A.  Newton,  to  whom  I  am  indebted  for  numerous  and  im 
portant  suggestions. 


CONTENTS, 


CHAPTER  I. 

DEFINITIONS   AND  NOTATION.  f^m 

General  Definitions 9 

Symbols  which  denote  Quantities 10 

Symbols  which  indicate  Operations 11 

Symbols  which  indicate  Relation 14 

Combination  of  Algebraic  Quantities 16 

CHAPTER  II. 

ADDITION 21 

CHAPTER  III. 

SUBTRACTION 25 

CHAPTER  IV. 

MULTIPLICATION. 

Case  of  Monomials — Case  of  Polynomials 32 

General  Theorems  proved 38 

CHAPTER  V. 

DIVISION. 

Case  of  Monomials — Case  of  Polynomials 41 

When  an±6n  is  exactly  divisible 47 

To  resolve  a  Polynomial  into  Factors 50 

CHAPTER  VI. 

GREATEST   COMMON   DIVISOR. — LEAST   COMMON  MULTIPLE. 

How  the  Greatest  Common  Divisor  is  found 52 

Rule  applied  to  Polynomials 55 

How  the  Least  Common  Multiple  is  found 58 

CHAPTER  VII. 

FRACTIONS. 

Definitions  and  General  Principles 62 

Reduction  of  Fractions 64 

Addition  and  Subtraction  of  Fractions...., 69 

Multiplication  and  Division  of  Fractions 72 


VI  CONTENTS. 

CHAPTER  VIII. 

EQUATIONS    OF    THE    FIRST   DEGREE. 

Definitions — Axioms  employed 79 

Transformation  of  Equations 81 

Solution  of  Equations 84 

Solution  of  Problems 87 

CHAPTER  IX. 

EQUATIONS   WITH   MORE   THAN   ONE    UNKNOWN  QUANTITY. 

Different  Methods  of  Elimination 99 

Equations  containing  three  or  more  Unknown  Quantities 105 

Problems  involving  several  Unknown  Quantities 109 

CHAPTER  X. 

DISCUSSION   OF   PROBLEMS. 

Positive  Values  of  x — Negative  Values 114 

Zero  and  Infinity 118 

Indeterminate  Solutions 121 

Inequalities 122 

CHAPTER  XL 

INVOLUTION. 

Powers  of  Monomials 127 

Powers  of  Polynomials 130 

CHAPTER  XII. 

EVOLUTION. 

Roots  of  Monomials 133 

Square  Root  of  Polynomials 136 

Square  Root  of  Numbers 139 

Cube  Root  of  Polynomials 143 

Cube  Root  of  Numbers 146 

CHAPTER  XIIL 

RADICAL    QUANTITIES. 

Transformation  of  Radical  Quantities , 150 

Addition  and  Subtraction  of  Radical  Quantities 157 

Multiplication  and  Division  of  Radical  Quantities 159 

Powers  and  Roots  of  Radical  Quantities 164 

Operations  on  Imaginary  Quantities 166 

To  find  Multipliers  which  shall  render  Surds  rational 168 

Square  Root  of  a  Binomial  Surd 171 

Equations  containing  Radical  Quantities 174 


PAGB 


CONTENTS.  Vii 


CHAPTER  XIV. 

EQUATIONS    OF    THE    SECOND   DEGREE. 

Incomplete  Equations  of  the  Second  Degree 177 

Complete  Equations  of  the  Second  Degree 182 

Equations  which  may  be  solved  like  Quadratics 187 

Problems  producing  Equations  of  the  Second  Degree 192 

Equations  containing  two  unknown  Quantities 196 

General  Properties  of  Equations  of  the  Second  Degree 203 

Discussion  of  the  general  Equation  of  the  Second  Degree 206 

Discussion  of  particular  Problems 209 

Geometrical  Construction  of  Equations 211 


CHAPTER  XV. 

RATIO   AND   PROPORTION. 

Definitions  ,.± 214 

Properties  of  Proportions 216 

Variation 222 

Problems  in  Proportion 224 

CHAPTER  XVI. 

PROGRESSIONS. 

Arithmetical  Progression — General  Principles 227 

Examples 230 

Geometrical  Progression — General  Principles 233 

Examples 237 

CHAPTER  XVII. 

CONTINUED   FRACTIONS. — PERMUTATIONS   AND   COMBINATIONS. 

Properties  of  Continued  Fractions 240 

Permutations  and  Combinations ,  244 

CHAPTER  XVIII. 

BINOMIAL     THEOREM. 

Powers  of  a  Binomial 248 

Sir  Isaac  Newton's  Binomial  Theorem 251 

Powers  and  Roots  of  Polynomials 255 

To  extract  any  Root  of  a  Number 257 

CHAPTER  XIX. 

SERIES. 

Definitions — Method  of  Differences 259 

Interpolation 263 

Development  of  Algebraic  Expressions  into  Series 265 


Vlll  CONTENTS. 

Method  of  Undetermined  Coefficients 267 

To  resolve  a  Fraction  into  Simpler  Fractions 272 

Reversion  of  Series 273 

General  Demonstration  of  the  Binomial  Theorem 275 

Applications  of  the  Binomial  Theorem 278 

CHAPTER  XX. 

LOGARITHMS. 

Properties  of  Logarithms 284 

The  common  System  of  Logarithms 287 

Multiplication  and  Division  by  Logarithms 293 

Involution  and  Evolution  by  Logarithms 294 

Exponential  Equations 296 

Compound  Interest  and  Annuities 297 

Computation  of  Logarithms 301 

CHAPTER  XXI. 

GENERAL  THEORY  OF  EQUATIONS. 

Definitions  and  general  Properties 306 

An  Equation  of  the  nth  Degree  has  n  Roots 308 

Law  of  the  Coefficients  of  every  Equation 310 

Number  of  imaginary  Roots 315 

Transformation  of  Equations 316 

Descartes's  Rule  of  Signs 320 

Composition  of  derived  Polynomials 321 

Equal  Roots , 322 

Sturm's  Theorem 323 

Solution  of  simultaneous  Equations  of  any  Degree 331 

CHAPTER  XXH. 

NUMERICAL   EQUATIONS   OP   HIGHER  DEGREES. 

Commensurable  Roots 334 

Incommensurable  Roots — Horner's  Method 338 

Equations  of  the  fourth  and  higher  Degrees 347 

Newton's  Method  of  Approximation 350 

Method  of  double  Position 351 

MISCELLANEOUS  EXAMPLES 355 


A  L  G  E  B 


CHAPTEE  I. 

DEFINITIONS  AND  NOTATION. 

1.  Quantity  is  any  thing  that  can  be  increased  or  diminished, 
and  that  can  be  measured. 

A  line,  a  surface,  a  solid,  a  weight,  etc.,  are  quantities;  but 
the  operations  of  the  mind,  such  as  memory,  imagination,  judg 
ment,  etc.  r *are  not  quantities. 

A  quantity  is  measured  by  finding  how  many  times  it  con 
tains  some  other  quantity  of  the  same  kind  taken  as  a  stand 
ard.  The  assumed  standard  is  called  the  unit  of  measure. 

2.  Mathematics  is  the  science  of  quantity,  or  the  science  which 
treats  of  the  properties  and  relations  of  quantities.     It  employs 
a  variety  of  symbols  to  express  the  values  and  relations  of 
quantities,  and  the  operations  to  be  performed  upon  these  quan 
tities,  or  upon  the  numbers  which  represent  these  quantities. 

3.  Mathematics  is  divided  into  pure  and  mixed.     Pure  math 
ematics  comprehends  all  inquiries  into  the  relations  of  quan 
tity  in  the  abstract,  and  without  reference  to  material  bodies. 
It  embraces  numerous  subdivisions,  such  as  Arithmetic,  Al 
gebra,  Geometry,  etc. 

In  the  mixed  mathematics,  these  abstract  principles  are  ap 
plied  to  various  questions  which  occur  in  nature.  Thus,  in  Sur 
veying,  the  abstract  principles  of  Geometry  are  applied  to  the 
measurement  of  land ;  in  Navigation,  the  same  principles  are 
applied  to  the  determination  of  a  ship's  place  at  sea ;  in  Optics, 
they  are  employed  to  investigate  the  properties  of  light;  and 
in  Astronomy,  to  determine  the  distances  of  the  heavenly 

bodies. 

A  2 


10  ALGEBKA. 

4.  Algebra  is  that  branch  of  mathematics  in  which  quantities 
are  represented  by  letters,  and  their  relations  to  each  other,  as 
well  as  the  operations  to  be  performed  upon  them,  are  indi- 
catfjc},  by  sigtoS  *cfr»:'&y  rebels.     The  object  of  algebraic  notation 
js.tp  abridge  ^nd  generalize  the  reasoning  employed  in  the  so- 
Tu!t.feri.rf^rqi(estiortS''.Tel.ating  to  numbers.    Algebra  may  there 
fore  be  called  a  species  of  Universal  Arithmetic. 

5.  The  symbols  employed  in  Algebra  may  be  divided  into 
three  classes : 

1st.  Symbols  which  denote  quantities. 

2d.  Symbols  which  indicate  operations  to  be  performed  upon 
quantities. 

3d.  Symbols  which  indicate  the  relations  subsisting  between 
different  quantities,  with  respect  to  their  magnitudes,  etc. 

Symbols  which  denote  Quantities. 

6.  In  order  to  generalize  our  reasoning  respecting  numbers, 
we  represent  them  by  letters,  as  a,  &,  c,  or  cc,  y,  z,  etc.,  and  these 
may  represent  any  numbers  whatever.     The  quantities  thus 
represented  may  be  either  known  quantities — that  is,  quantities 
whose  values  are  given ;  or  unknown  quantities — that  is,  quan 
tities  whose  values  are  to  be  determined. 

Known  quantities  are  generally  represented  by  the  first  let 
ters  of  the  alphabet,  as  a,  6,  c,  d,  etc.,  and  unknown  quantities  by 
the  last  letters  of  the  alphabet,  as  jc,  */,  z,  u,  etc.  This,  how 
ever,  is  not  a  necessary  rule,  and  is  not  always  observed. 

7o  Sometimes  several  quantities  are  represented  by  a  single 
letter,  repeated  with  different  accents,  as  a',  a",  a!" ,  a"" ,  etc., 
which  are  re£d  a  prime,  a  second,  a  third,  etc. ;  or  by  a  letter 
repeated  with  different  subscript  figures,  as  av  a2,  «3,  «4,  etc., 
which  may  be  read  a  one  sub,  a  two  sub,  a  three  sub,  etc. 
All  these  symbols  represent  different  quantities,  but  the  ac 
cents  or  numerals  are  employed  to  indicate  some  important  re 
lation  between  the  quantities  represented. 


DEFINITIONS   AND   NOTATION.  11 

8.  Sometimes  quantities  are  represented  by  the  initial  letters 
of  their  names.     Thus  s  may  represent  sum;  d,  difference  or  di 
ameter  ;  r,  radius  or  ratio ;  c,  circumference ;  h,  height,  etc.     All 
these  letters  may  be  used  with  accents.     Thus,  in  a  problem, 
relating  to  two  circles,  d  may  represent  the  diameter  of  one  cir 
cle,  and  df  the  diameter  of  the  other;  c  the  circumference  of 
one,  and  c'  the  circumference  of  the  other,  etc. 

Symbols  which  indicate  Operations. 

9.  The  sign  of  addition  is  an  erect  cross,  +  ,  called  plus,  and 
when  placed  between  two  quantities  it  indicates  that  the  second 
is  to  be  added  to  the  first.     Thus,  5  +  3  indicates  that  we  must 
add  3  to  the  number  5,  in  which  case  the  result  is  8.     We  also 
make  use  of  the  same  sign  to  connect  several  numbers  togeth 
er.     Thus,  7  +  5  +  9  indicates  that  to  the  number  7  we  must 
add  5  and  also  9,  which  make  21.     So,  also,  8  +  5  +  13  +  11  +  1 
+  3  +  10  is  equal  to  51. 

The  expression  a-\-b  indicates  the  sum  of  two  numbers,  which 
we  represent  by  a  and  b.  In  the  same  manner,  m-\-n-\-x-\-y 
indicates  the  sum  of  the  numbers  represented  \)j  these  four  let 
ters.  If  we  knew,  therefore,  the  numbers  represented  by  the 
letters,  we  could  easily  find  by  arithmetic  the  value  of  such  ex 
pressions. 

10.  The  sign  of  subtraction  is  a  short  horizontal  line,  — ,  called 
minus.     When  placed  between  two  quantities,  it  indicates  that 
the  second  is  to  be  subtracted  from  the  first.     Thus,  8  —  5  indi 
cates  that  the  number  5  is  to  be  taken  from  the  number  8, 
which  leaves  a  remainder  of  3.     In  like  manner,  12  —  7  is  equal 
to  5,  etc. 

Sometimes  we  may  have  several  numbers  Jto  subtract  from  a 
single  one.  Thus,  16  —  5  —  4  indicates  that  5  is  to  be  subtracted 
from  16,  and  this  remainder  is  to  be  further  diminished  by  4, 
leaving  7  for  the  result.  In  the  same  manner,  50  —  1  —  5  —  3  — 
9  — 7  is  equal  to  25. 

The  expression  a  —  b  indicates  that  the  number  designated  by 
c  is  to  be  diminished  by  the  number  designated  by  b. 


12  ALGEBRA. 

11  „  The  double  sign  ±  is  sometimes  written  before  a  quan 
tity  to  indicate  that  in  certain  cases  it  is  to  be  added,  and  in 
others  it  is  to  be  subtracted.  Thus,  b  ±  c  is  read  b  plus  or 
minus  c,  and  denotes  either  the  sum  or  the  difference  of  these 
two  quantities. 

12.  The  sign  of  multiplication  is  an  inclined  cross,  x.    When 
placed  between  two  quantities,  it  indicates  that  the  first  is  to 
be  multiplied  by  the  second.     Thus,  3x5  indicates  that  3  is 
to  be  multiplied  by  5,  making  15.     In  like  manner,  axb  indi 
cates  that  a  is  to  be  multiplied  by  b;   and  axbxc  indicates 
the  continued  product  of  the  numbers  designated  by  a,  5,  and 
c,  and  so  on  for  any  number  of  quantities. 

Multiplication  is  also  frequently  indicated  by  placing  a  point 
between  the  successive  letters.  Thus,  a.b.c.d  signifies  the 
same  thing  as  axbxcxd. 

Generally,  however,  when  numbers  are  represented  by  let 
ters,  their  multiplication  is  indicated  by  writing  them  in  suc 
cession  without  any  intervening  sign.  Thus,  abc  signifies 
the  same  as  axbxc,  or  a.b.c. 

The  notation  a.b  or  ab  is  seldom  employed  except  when 
the  numbers  are  designated  by  letters.  If,  for  example,  we 
attempt  to  represent  in  this  manner  the  product  of  the  num 
bers  5  and  6,  5.6  might  be  confounded  with  5^;  and  56 
would  be  read  fifty-six,  instead  of  five  times  six. 

The  multiplication  of  numbers  may,  however,  be  denoted 
by  placing  a  point  between  them  in  cases  where  no  ambiguity 
can  arise  from  the  use  of  this  symbol.  Thus,  1.2.3.4.5  is 
sometimes  used  to  represent  the  continued  product  of  the 
numbers  1,  2,  3,  4,  5. 

13.  When  two  or  more  quantities  are  multiplied  together, 
each  of  them  is  called  &  factor.     Thus,  in  the  expression  7x5, 
7  is  a  factor,  and  so  is  5.     In  the  product  abc  there  are  three 
factors,  a,  &,  c. 

When  a  quantity  is  represented  by  a  letter,  it  is  called  a 
literal  factor.  When  it  is  represented  by  a  figure  or  figures,  it 


DEFINITIONS   AND   NOTATION.  13 

is  called  a  numerical  factor.     Thus,  in  the  expression  5ab,  5 
is  a  numerical  factor,  while  a  and  b  are  literal  factors. 

14.  The  sign  of  division  is  a  short  horizontal  line  with  a 
point  above  and  one  below,  -=-.  When  placed  between  two 
quantities,  it  indicates  that  the  first  is  to  be  divided  by  the 
second. 

Thus,  24-^-6  indicates  that  24  is  to  be  divided  by  6,  making 
4.  So,  also,  a-^b  indicates  that  a  is  to  be  divided  by  b. 

Generally,  however,  the  division  of  two  numbers  is  indi 
cated  by  writing  the  divisor  under  the  dividend,  and  drawing 
a  line  between  them.  Thus,  24-^-6  and  a  H-6  are  usually  writ- 

ten        and    . 


15.  The  products  formed  by  the  successive  multiplication 
of  the  same  number  by  itself,  are  called  the  powers  of  that 
number. 

Thus,  2  x2=4,  the  second  power  or  square  of  2. 
2x2x2  =  8,  the  third  power  or  cube  of  2. 

2  x  2  x  2  x  2  =  16,  the  fourth  power  of  2,  etc. 
So,  also,  3x3  =  9,  the  second  power  of  3. 

3  x  3  X  3  =  27,  the  third  power  of  3,  etc. 
Also,       a  x  a  =  aa,  the  second  power  of  a. 

ax  ax  a  =  aaa,  the  third  power  of  a,  etc. 
In  general,  any  power  of  a  quantity  is  designated  by  the 
number  of  equal  factors  which  form  the  product. 

16.  The  sign  of  involution  is  a  number  written  above  a  quan 
tity,  at  the  right  hand,  to  indicate  how  many  times  the  quan 
tity  is  to  be  taken  as  a  factor. 

A  root  of  a  quantity  is  a  factor  which,  multiplied  by  itself 
a  certain  number  of  times,  will  produce  the  given  quantity. 

The  figure  which  indicates  how  many  times  the  root  or  fac 
tor  is  taken,  is  called  the  exponent  of  the  power. 

Thus,  instead  of  aa,  we  write  a2,  where  2  is  the  exponent  of 
the  power;  instead  of  aaa,  we  write  a3,  where  3  is  the  expo- 


14  ALGEBRA. 

nent  of  the  power ;   instead  of  aaaaa,  we  write  a5,  where  5  is 
the  exponent  of  the  power,  etc. 

When  no  exponent  is  written  over  a  quantity,  the  exponent 

1  is  always  understood.    Thus,  a1  and  a  signify  the  same  thing. 

Exponents  may  be  attached  to  figures  as   well  as  letters. 

Thus  the  product  of  3  by  3  may  be  written  32,  which  equals  9. 

"      3x3x3      "     33,    "     27, 

"      3x3x3x3x3"      35,     "    243. 

17.  The  sign  of  evolution,  or  the  radical  sign,  is  the  charac 
ter  <|/~.    When  placed  over  a  quantity,  it  indicates  that  a  root 
of  that  quantity  is  to  be  extracted.     The  name  or  index  of  the 
required  root  is  the  number  written  above  the  radical  sign. 
Thus, 

V/9,  or  simply  V9,  denotes  the  square  root  of  9,  which  is  3. 

V/64  denotes  the  cube  root  of  64,  which  is  4. 

Vl6  denotes  the  fourth  root  of  16,  which  is  2. 
So,  also, 

I/a,  or  simply  -y/a,  denotes  the  square  root  of  a. 

I/a  denotes  the  fourth  root  of  a. 

I/a  denotes  the  nth.  root  of  a,  where  n  may  represent  any 
number  whatever. 

When  no  index  is  written  over  the  sign,  the  index  2  is  un 
derstood.  Thus,  instead  of  Vab,  we  usually  write  Vab. 

Symbols  ivliich  indicate  Relation. 

18.  The  sign  of  equality  consists  of  two  short  horizontal 
lines,  =.     When  written  between  two  quantities,  it  indicates 
that  they  are  equal  to  each  other. 

Thus,  7  +  6  =  13  denotes  that  the  sum  of  7  and  6  is  equal  to 
13. 

In  like  manner,  a  =  b  +  c  denotes  that  a  is  equal  to  the  sum 
of  b  and  c ;  and  a+b=c— d  denotes  that  the  sum  of  the  num 
bers  designated  by  a  and  5,  is  equal  to  the  difference  of  the 
numbers  designated  by  c  and  d. 


DEFINITIONS   AND    NOTATION.  15 

19.  The  sign  of  inequality  is  the  angle  >   or  <.     When 
placed  between  two  quantities,  it  indicates  that  they  are  un 
equal,  the  opening  of  the  angle  being  turned  toward  the  greater 
number.     When  the  opening  is  toward  the  left,  it  is  read  great 
er  than  ;  when  the  opening  is  toward  the  right,  it  is  read  less 
than.     Thus,  5>3  denotes  that  5  is  greater  than  3  ;  and  6<11 
denotes  that  6  is  less  than  11.     So,  also,  a>6  denotes  that  a  is 
greater  than  b;  and  x<y+z  denotes  that  x  is  less  than  the 
sum  of  y  and  z. 

20.  A  parenthesis,  (  ),  or  a  vinculum,  -  ,  is  employed  to 
connect  several  quantities,  all  of  which  are  to  be  subjected  to 
the  same  operation.  _ 


_ 

Thus  the  expression  (a-f-5-fc)xa?,  or  «  +  />  +  cxx,  indicates 
that  the  sum  of  a,  6,  and  c  is  to  be  multiplied  by  x.  But 
a  +  b+cxoc  denotes  that  c  only  is  to  be  multiplied  by  x. 

When  the  parenthesis  is  used,  the  sign  of  multiplication 
is  generally  omitted.  Thus,  (a  +  Z>-fc)xx'  is  the  same  as 


21.  The  sign  of  ratio  consists  of  two  points  like  the  colon  : 
placed  between  the  quantities  compared.     Thus  the  ratio  of  a 
to  b  is  written  a  :  b. 

22.  The  sign  of  proportion  consists  of  a  combination  of  the 
sign  of  ratio  and  the  sign  of  equality,  thus,  :  =  :  ;  or  a  com 
bination  of  eight  points,  thus,  :   ::   :. 

Thus,  if  a,  &,  c,  d,  are  four  quantities  which  are  proportional 
to  each  other,  we  say  a  is  to  b  as  c  is  to  d;  and  this  is  express 
ed  by  writing  them  thus: 

a:  b  =  c  :  c?, 
or  a  :  b  :  :  c  :  d. 

23.  The  sign  of  variation  is  the  character  <x>.     When  written 
between  two  quantities,  it  denotes  that  both  increase  or  diminish 
together,  and  in  the  same  ratio.    Thus  the  expression  sootv  de 
notes  that  s  varies  in  the  same  ratio  as  the  product  of  t  and  v. 


16  ALGEBRA. 

24.  Three  dots  /.  are  sometimes  employed  to  denote  there' 
fore,  or  consequently. 

A  few  other  symbols  are  employed  in  Algebra,  in  addition  to 
those  here  enumerated,  which  will  be  explained  as  they  occur. 

Combination  of  Algebraic  Quantities. 

25,  Every  number  written  in  algebraic  language — that  is, 
by  aid  of  algebraic  symbols — is  called  an  algebraic  quantity,  or 
an  algebraic  expression. 

Thus,  3a2  is  the  algebraic  expression  for  three  times  the 
square  of  the  number  a. 

7a364  is  the  algebraic  expression  for  seven  times  the  third 
power  of  a,  multiplied  by  the  fourth  power  of  b. 

28.  An  algebraic  quantity,  not  composed  of  parts  which  are 
separated  from  each  other  by  the  sign  of  addition  or  subtrac 
tion,  is  called  a  monomial,  or  a  quantity  of  one  term,  or  simply 
a  term. 

Thus,  3a,  55c,  and  7a??/2,  are  monomials. 

Positive  terms  are  those  which  are  preceded  by  the  sign  plus, 
and  negative  terms  are  those  which  are  preceded  by  the  sign 
minus.  When  the  first  term  of  an  algebraic  quantity  is  posi 
tive,  the  sign  is  generally  omitted.  Thus  a  +  b  —  c  is  the  same 
as  -\-a  +  b  — c.  The  sign  of  a  negative  term  should  never  be 
omitted. 

27.  The  coefficient  of  a  quantity  is  the  number  or  letter  pre 
fixed  to  it,  showing  how  often  the  quantity  is  to  be  taken. 

Thus,  instead  of  writing  a  +  a  +  a  +  a+a,  which  represents 
5  a's  added  together,  we  write  5a,  where  5  is  the  coefficient  of 
a.  In  §(x+y\  6  is  the  coefficient  of  x-\-y.  When  no  coeffi 
cient  is  expressed,  1  is  always  to  be  understood.  Thus,  la  and 
a  denote  the  same  thing. 

The  coefficient  may  be  a  letter  as  well  as  a  figure.  In  the  ex 
pression  nx,  n  may  be  considered  as  the  coefficient  of  x,  be 
cause  x  is  to  be  taken  as  many  times  as  there  are  units  in  n. 
If  n  stands  for  5,  then  nx  is  5  times  x.  When  the  coefficient 


DEFINITIONS  AND  NOTATION.  17 

is  a  number,  it  may  be  called  a  numerical  coefficient  ;  and  when 
it  is  a  letter,  a  literal  coefficient. 

In  7ax,  7  may  be  regarded  as  the  coefficient  of  ax,  or  7a  may 
be  regarded  as  the  coefficient  of  x. 

28  The  coefficient  of  a  positive  term  shows  how  many  times 
the  quantity  is  taken  positively,  and  the  coefficient  of  a  nega 
tive  term  shows  how  many  times  the  quantity  is  taken  nega 
tively.  Thus,  -f  4x  =  +  x  +  x  +  x  +  x  ; 


29.  Similar  terms  are  terms  composed  of  the  same  letters, 
affected  with  the  same  exponents.     The  signs  and  coefficients 
may  differ,  and  the  terms  still  be  similar. 

Thus,  Sab  and  lab  are  similar  terms. 
Also,  5a2c  and  —  3a2c  are  similar  terms. 

30.  Dissimilar  terms  are  those  which  have  different  letters  or 
exponents. 

Thus,  axy  and  axz  are  dissimilar  terms. 
Also,  Bab2  and  4a26  are  dissimilar  terms. 

31.  A  polynomial  is  an  algebraic  expression  consisting  of 
more  than  one  term;  as,  a  +  b  ;  or  a  +  2b  —  5c  +  x. 

A  polynomial  consisting  of  two  terms  only  is  usually  called 
a  binomial;  and  one  consisting  of  three  terms  only  is  called  a 
trinomial.  Thus,  3a  +  5&  is  a  binomial;  and  5a—3bc-\-xy  is  a 
trinomial. 

32.  The  degree  of  a  term  is  the  number  of  its  literal  factors. 
Thus,  3a  is  a  term  of  the  first  degree. 

Sab  "  second  " 

6a25c3         "  sixth     " 

In  general,  the  degree  of  a  term  is  found  by  taking  the  sum 
of  the  exponents  of  all  the  letters  contained  in  the  term. 

Thus  the  degree  of  the  term  5ab2cd3  is  1  +  2  +  1  -f  3,  or  7; 
that  is,  this  term  is  of  the  seventh  degree. 


18  ALGEBRA. 

33.  A  polynomial  is  said  to  be  homogeneous  when   all  its 
terms  are  of  the  same  degree. 
Thus,  3a2—  4ab-}-b2  is  of  the  second  degree,  and  homogeneous. 

2a3-f3a2c-4c2d     "     third          "  " 

But     5a3  —  2ab  +  c  is  not  homoeneous. 


34.  The  reciprocal  of  a  quantity  is  the  quotient  arising  from 
dividing  a  unit  by  that  quantity. 

Thus  the  reciprocal  of  2  is  J;   the  reciprocal  of  a  is  £ 

35.  A  function  of  a  quantity  is  any  expression  containing 
that  quantity.     Thus, 

is  a  function  of  a?. 

is  a  function  of  y. 
ax2—  by2  is  a  function  of  x  and  y. 

Exercises  in  Algebraic  Notation. 

36.  In  the  following  examples  the  pupil  is  simply  required 
to  express  given  relations  in  algebraic  language. 

Ex.  1.  Give  the  algebraic  expression  for  the  following  state 
ment:  The  second  power  of  a,  increased  by  twice  the  product 
of  a  and  6,  diminished  by  c,  and  increased  by  c?,  is  equal  to 
fifteen  times  x.  Ans.  «2  +  2a&  —  c  -\-d-l5x. 

Ex.  2.  The  quotient  of  three  divided  by  the  sum  of  x  and 
four,  is  equal  to  twice  b  diminished  by  eight. 

Ex.  3.  One  third  of  the  difference  between  six  times  x  and 
four,  is  equal  to  the  quotient  of  five  divided  by  the  sum  of  a 
and  b. 

Ex.  4.  Three  quarters  of  x  increased  by  five,  is  equal  to 
three  sevenths  of  b  diminished  by  seventeen. 

Ex.  5.  One  ninth  of  the  sum  of  six  times  x  and  five,  added 
to  one  third  of  the  sum  of  twice  x  and  four,  is  equal  to  the 
product  of  a,  5,  and  c. 

Ex.  6.  The  quotient  arising  from  dividing  the  sum  of  a  and 
I  by  the  product  of  c  and  c?,  is  greater  than  four  times  the  sum 
of  m:  n,  x,  and  y. 


DEFINITIONS  AND    NOTATION.  19 

37.  In  the  following  examples  the  pupil  is  required  to  trans 
late  the  algebraic  symbols  into  common  language. 


x       m 
1  —  =• 


b        c     a  -f  b' 

Ans.  The  quotient  arising  from  dividing  the  sum  of  a  and 
x  by  &,  increased  by  the  quotient  of  x  divided  by  c,  is  equal  to 
the  quotient  of  m  divided  by  the  sum  of  a  and  b. 

Ex.  2.  7a2-\-(b — c)  X  (d-\-e)=x-\-y. 

How  should  the  preceding  example  be  read  when  the  first 
parenthesis  is  omitted? 


Ex.4,. 

8«  —  d 

Ex.  5.  2a  V^?  —  ac  =  6(a  +  m+x). 
v 


Computation  of  Numerical  Values. 

38.  The  numerical  value  of  an  algebraic  expression  is  the  re 
sult  obtained  when  we  assign  particular  values  to  all  the  let 
ters,  and  perform  the  operations  indicated. 

Suppose  the  expression  is  2azb. 

If  we  make  a  =  2  and  5  =  3,  the  value  of  this  expression  will 
be  2x2x2x3  =  24. 

If  we  make  a=4  and  5  =  3,  the  value  of  the  same  expression 
will  be  2x4x4x3  =  96. 

The  numerical  value  of  a  polynomial  is  not  affected  by 
changing  the  order  of  the  terms,  provided  we  preserve  their  re 
spective  signs. 

The  expressions  a2  +  2a/;  +  62,  az  +  l2  +  2ab,  and  62-f  2«&  +  a2, 
have  all  the  same  numerical  value. 

Find  the  numerical  values  of  the  following  expressions,  in 
which  a  =  6,  6  =  5,  c  =  4,  ra  =  8,  and  n-2. 

Ex.l.  a2+3«6-c2.  Ans.  36  +  90-16  =  110. 

Ex.  2.  a2x(a-h&)-2a£c.  Ans.  156. 


20 


ALGEBRA, 


c.  5.   -v/62  — 

.  6.  3v'c+2aV2a-f-£+2c. 
.  7.  (3  A/C+  2a)  V2a 


—  n     ra  +  n 


JLx.  9.  - 

Find  the  numerical  values  of  the  following  expressions,  in 
which  a=3,  6=5,  c=2,  m=4,  n=6,  and  x=:9. 


.  10. 


x  —  or—  c 

Ex.  11.  5,r—  7  Vos. 
JB5p.  12.  2x2 
JEfc.13.  VlO+w-VlO+n. 

14.  5(m2  +  ?i2)+4ax. 
.  15.  a4-4a3+7a2-6a. 

.  16.   V5  VTW  +  S  V^H-  Vm+ 


Ans.  8. 


.  175. 


.. 

T/i  —  n-\-o 


n 


Ex.  19.  —  o 

^ 


ADDITION.  2J 


CHAPTEK  II. 

ADDITION. 

39.  Addition,  in  Algebra,  is  the  connecting  of  quantities  to 
gether  by  means  of  their  proper  signs,  and  incorporating  such 
as  can  be  united  into  one  sum. 

When  the  Quantities  are  similar  and  have  the  same  Signs. 

40.  The  sum  of  3a,  4a,  and  5a,  is  obviously  12a.     That  is, 


So,  also,'— 3a,  —  4a,  and  — 5a,  make  —  12a;  for  the  minus 
sign  before  each  of  the  terms  shows  that  they  are  to  be  sub 
tracted,  not  from  each  other,  but  from  some  quantity  which  is 
not  here  expressed ;  and  if  3a,  4a,  and  5a  are  to  be  subtracted 
successively  from  the  same  quantity,  it  is  the  same  as  subtract 
ing  at  once  12a.  Hence  we  deduce  the  following 

RULE. 

Add  the  coefficients  of  the  several  quantities  together,  and  to  their 
sum  annex  the  common  letter  or  letters,  prefixing  the  common  sign. 

EXAMPLES. 

3a  —  3a&  2£  +  3#  a— 2x2           2a+  y2 

5a  —  6a&  5&-f7ce  4a— Bx2 

7a  —  ab  b  +  Zx  Sa — 5x2 

a  —7ab  46+ 3#  7a—   x2 


The  pupil  must  continually  bear  in  mind  the  remark  of  Art. 
/26,  that,  when  no  sign  is  prefixed  to  a  quantity,  plus  is  always 
to  be  understood. 

When  the  Quantities  are  similar,  but  have  different  Signs. 

41.  The  expression  7a  —  4a  denotes  that  4a  is  to  be  sub* 
tracted  from  7a,  and  the  result  is  obviously  3a.  That  is, 

7a—  4a  =  3«. 


22  ALGEBRA. 

The  expression  5«  —  2a-f  3a  —  a  denotes  that  we  are  to  sub 
tract  2tt  from  5a,  add  3a  to  the  remainder,  and  then  subtract  a 
from  the  last  sum,  the  result  of  which  operation  is  5a.  That  is, 
5a— 2a+3a— a  —  oa. 

It  is  generally  most  convenient  to  take  the  sum  of  the  posi 
tive  quantities,  which  in  the  preceding  case  is  8a;  then  take 
the  sum  of  the  negative  quantities,  which  in  this  case  is  Sa  ;| 
and  we  have  8a— 3a,  or  5#,  the  same  result  as  before.  Hence 
we  deduce  the  following 

RULE. 

Add  all  the  positive  coefficients  together,  and  also  all  those  that 
are  negative;  subtract  the  least  of  these  results  from  the  greater; 
to  the  difference  annex  the  common  letter  or  letters,  and  prefix  the 
sign  of  the  greater  sum. 

EXAMPLES. 

-3a  6x+5ay  2ay-   7  -2a2x  _6a2  +  2& 

4.7^  -Sx  +  2ay  -  ay+  8  .          a2x  2a?-Sb 

_j_8a  x—6ay  2«y—   9  —  3a*x  .    —  5a2  —  Sb 

-_a  2:/:+   ay  Say -II  7a2x  4az-2b 

-4-1  la  (ix+Zay 

When  some  of  the  Quantities  are  dissimilar. 

42.  Dissimilar  terms  can  not  be  united  into  one  term  by  ad 
dition. 

Thus  2a  and  3&  neither  make  5a  nor  51).  Their  sum  can 
therefore  only  be  indicated  by  connecting  them  by  their  proper 
signs,  thus,  2a  +  3/>. 

In  adding  together  polynomials  which  contain  several  groups 
of  similar  quantities,  it  is  most  convenient  to  write  them  in 
such  a  manner  that  each  group  of  similar  quantities  may  occiv 
py  a  column  by  itself.  Hence  we  deduce  the  following 

RULE. 

Write  the  quantities  to  be  added  so  that  the  similar  terms  may 
be  arranged  in  the  same  column. 

Add  up  each  column  separately,  and  connect  the  several  results 
ly  their  proper  signs. 


ADDITION.  23 

EXAMPLES. 

1.  Add  2a+3Z>+4c,  ft+26  +  5c,  3a-6+2c,  an 

Ans. 

2.  Add  2xy-2x2  +  ?/2,  3x2-f  X7/+4?/2,  x2—  x?/4-3z/2,  and  4x2 
-2y2—  Sxy.  Ans.  6x2—  x?/  +  6?/2. 

3.  Add  5a2x2  —  2xy,  3ax  —  4x?/,  7x?/  —  4«x,  a2x2  +  5x?/,  and 
2ax—  3xy.  Aws.  6a2x2  4-  3x?/  -f  ##• 

4.  Add   2a2  —  3ac  +  36  —  ccZ,    4a2  —  ac  +  2cd—b,    3a2  +  2«c—  4i 
+  3c^,  and  a2-2«c  +  5c^-26.         ^4?is.  10a2-4ac-4&  +  9«Z. 

5.  Add    7w-f3rc  —  14x,    3aH-9w  —  llm,    5:r  —  4m  +  8w,    and 


6.  Add  2a 

and  3a2—  2^2x-f  2ax2.  Ans.  3a2x—  5x2+2rfx. 

7.  Add   2a262  —  3ax  +  5m2?/,    2w2?/  +  3a2&2-2ax,    4ax-3m2y 
—  4a2i2,  and  «x  +  3a262  —  tetfy.  Ans.  ±a2l2. 

8.  Add  7«£3-12ax2,   I3ab3  +  ax,   Sax?  -Sab3,   and    ~12«&3 
4-9ax2-5. 

9.  Add  4x2+2ax+l,  3ax-2x2  +  5?  3x2-6ax+4,  and  5x2 


10.  Add   2a3x2  —  3a2x3  —  a 

-|-  2ax2  +  3  ax,  a  n  d  3  a?x2  —  ax2  —  a2x3  +  4a2x  +  ax. 

11.  Add   14a3x  -  7a2i2  +  3a2,    5aW  +  3a2^2  +  2a2 
-5«3x-a2,  and  4a2//-9«3x-4a2. 

12.  Add   ax4- 

,  and  2 


43.  It  must  be  observed  that  the  term  addition  is  used  in  a 
more  general  sense  in  Algebra  than  in  Arithmetic.  In  Arith 
metic,  where  all  quantities  are  regarded  as  positive,  addition 
implies  augmentation.  The  sum  of  two  quantities  will  there 
fore  be  numerically  greater  than  either  quantity.  Thus  the 
sum  of  7  and  6  is  12,  which  is  numerically  greater  than  either 
5  or  7. 

But  in  Algebra,  the  quantities  to  be  added  may  be  either 
positive  or  negative;  and  by  the  sum  of  two  quantities  we  un 
derstand  their  aggregate,  taken  with  reference  to  their  signs. 


24  ALGEBRA. 

Thus  the  sum  of  +7  and  —5  is  +  2,  which  is  numerically  less 
than  either  7  or  5.  So,  also,  the  sum  of  +a  and  —  b  is  a  —b. 
In  this  case  the  algebraic  sum  "is  numerically  the  difference  of 
the  two  quantities. 

This  is  one  instance  among  many  in  which  the  same  terms 
are  used  in  a  much  more  general  sense  in  the  higher  mathe 
matics  than  they  are  in  Arithmetic. 

44.  When  dissimilar  terms  have  a  common  literal  part,  we 
may  regard  the  other  factors  as  the  coefficient  of  the  common 
letter  or  letters.  The  sum  of  the  terms  will  then  be  expressed 
by  inclosing  the  sum  of  the  coefficients  in  a  parenthesis,  and 
prefixing  it  to  the  common  letter  or  letters. 

Thus  the  sum  of  ax2,  bx2,  and  ex2  may  be  written 


EXAMPLES. 

1.  Add  ax,  2bx,  and  3rax.  Ans.  (a+2b+&m)x. 

2.  Add  Sax?/2,  2bxy2,  and  —  5axy2.  Ans.  (2b—  2a)xy2. 

3.  Add  2ax+3?/,  5ax—  ?/,  and  x—4y.      Ans.  (7a+l)x—2y. 

4.  Add  2x-f  3xy,  ax  +  te/,  and  bx+3mxy. 

5.  Add  mx  +  ny,  3ax—%y,  and  4bx+ay. 

6.  Add  4raV^  +  3,  2aV^—  1,  and  bi/x  +  y. 

7.  Add  3ax2  +  2&x—  1,  45x2  —  ax  +  3,  and  rax2—  nx  +  5. 

8.  Add  2ax4-f  36x3  —  7,  3mx4-nx3  +  2,  and  4x4—  ax3  +  l. 

9.  Add  awx3-fZwx2-hcx,  bmx3—  a??x2  +  ax,  and   cmx3—nxr 


10.  Add  (a-b)Vx  and 


SUBTiiACTiON.  25 


CHAPTER  III. 

SUBTRACTION. 

45.  Subtraction  is  the  operation  of  finding  the  difference  be 
tween  two  quantities  or  sets  of  quantities.  The  quantity  to  be 
subtracted  is  called  the  subtrahend ;  the  quantity  from  which 
it  is  to  be  subtracted  is  called  the  minuend;  the  quantity 
which  is  left  after  the  subtraction  is  called  the  remainder. 

Let  it  be  required  to  subtract  8  —  3  from  15. 

Now  8  —  S  is  equal  to  5;  and  5  subtracted  from  15  leaves 
10.  The  result,  then,  must  be  10.  But,  to  perform  the  opera 
tion  on  the  numbers  as  they  were  given,  we  first  subtract  8 
from  15,  and  obtain  7.  This  result  is  too  small  by  3,  because 
the  number  8  is  larger  by  3  than  the  number  which  was  re 
quired  to  be  subtracted,  Therefore,  in  order  to  correct  this  re 
sult,  the  3  must  be  added,  and  the  operation  may  be  expressed 

thua'  15-8  +  3  =  10. 

Again,  let  it  be  required  to  subtract  c-^-d  from  a— b.  It  is 
plain  that,  if  the  part  c  were  alone  to  be  subtracted,  the  re 
mainder  would  be  7 

Ct  "~~  0  — —  C. 

But,  since  the  quantity  actually  proposed  to  be  subtracted  is 
less  than  c  by  d,  too  much  has  been  taken  away  by  c?,  and  there 
fore  the  true  remainder  will  be  greater  than  a—b—c  by  c?,  and 
may  hence  be  expressed  thus, 

a  —  b^~ 

where  the  signs  of  the  last  two  terms  are  both  contrary  to  what 
they  were  given  in  the  subtrahend.  Hence  we  perceive  that 
a  quantity  is  subtracted  by  simply  changing  its  sign.  In  prac 
tice  it  is  most  convenient  to  write  the  quantities  so  that  simi 
lar  terms  may  be  found  in  the  same  column. 

B 


26  ALGEBRA. 

46.  Hence  we  deduce  the  following 

RULE. 

Write  the  subtrahend  under  the  minuend,  arranging  similar 
terms  in  the  same  column. 

Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to  be 
changed  from  -f  to  —  ,  or  from  —  to  -f  ,  and  then  proceed  as  in 
addition. 

EXAMPLES. 

1.  2.  3.  4.                     5. 

From            I5ax2  5abx2  4abx3  —  Babx2  —  I2a2bc 

Subtract         Tax2  llabx2  —  Sabx3  —  Sabx2             8a2bc 

Eemainder     Sax2  —ftabx2  7abx3  5abx2  — 


6.  7.  8. 

From  3a2  -\-4b-2x  5x2-3ax2+ll 

Subtract         cP  +  tb-Sx  7x2-Sax2-   2         a?-2ab+b 


Remainder  2a2-36  +  6x         -2x2+5ax2  +  13  4ab 


9.  From  3a+5Z>—  2c  subtract  2a—  b.        Ans.  a  +  66—  2c. 

10.  From  5aic—  2&—  6  subtract  3aic—  2&+1. 

Ans.2abc—7. 

11.  From  4«2—  7a-h3x  subtract  a2+3a—  2x. 

Ans.  3«2—  10a  +  5x. 

12.  From  a  —  6-f2ra—  x  subtract  3x+m—  4b  +  a. 

13.  From  2x3—  x2y+5xy2  subtract  x3—2xy2+y3. 

14.  From  m  +  n  subtract  m  —  n. 

15.  From  m  -\-n-\-x  subtract  —m—n—x. 

16.  From  5a2  —  3a—  7  subtract  —  2a2—  4a+10. 

17.  From  ??i44-37?^3  —  4m2—  2m+l   subtract  m4—  2m3  +  m2 
—  3m  +  5. 

18.  From  x5-5x4+10x3-3  subtract  x5  +  5x4-10x3+3. 

19.  From  3a2+ax  +  2x2—  14a2x   subtract   x2  —  15a2x  -f  2a2 


20.  From  6a5x—  4m?i+5acc  subtract  3mn-}-6ax—  3a&x. 

Subtraction  may  be  proved,  as  in  Arithmetic,  by  adding  the 
remainder  to  the  subtrahend.  The  sum  should  be  equal  to 
the  minuend. 


SUBTRACTION.  27 

47.  It  will  be  perceived  that  the  term  subtraction  is  used  in 
a  more  general  sense  in  Algebra  than  in  Arithmetic.     In  Arith 
metic,  where  all  quantities  are  regarded  as  positive,  a  number 
is  always  diminished  by  subtraction.     But  in  Algebra  the  dif 
ference  between   two  quantities  may  be  numerically  greater 
than  either.     Thus  the  difference  between   -f  a  and  —  b  is 
a-\-b. 

The  distinction  between  positive  and  negative  quantities 
may  be  illustrated  by  the  scale  of  a  thermometer.  The  de 
grees  above  zero  are  considered  positive,  and  those  below  zero 
negative.  From  five  degrees  above  zero  to  five  degrees  below 
zero,  the  numbers  stand  thus, 

+  5,  d-4,  +3,  +  2,  +1,  0,  -1,  -2,  -3,  -4,  -5. 

The  difference  between  a  temperature  five  degrees  above 
zero  and  one  which  is  five  degrees  Mow  zero,  is  ten  degrees, 
which  is  numerically  the  sum  of  the  two  quantities.  Ten  is 
said  to  be  the  algebraic  difference  between  -}-5  and  —5. 

48.  When  dissimilar  terms  have  a  common  literal  part,  the 
difference  of  the  terms  may  be  expressed,  as  in  Art.  44,  by  in 
closing  the  difference  of  the  coefficients  in  a  parenthesis,  and 
prefixing  it  to  the  common  letter  or  letters. 

Thus  the  difference  between  ax2  and  bx2  may  be  written 
(a-b)x2. 

EXAMPLES. 

1.  From  ax2?/2  subtract  —  3x2if.  Ans.  (a  -\-  3)x2?/2. 

2.  From  2aa;  +  3?/  subtract  5bx—y.        Ans.  (2a— 

3.  From  mx-\-ny  subtract  Sax—2y. 

Ans.  (m  —  3a)x 

4.  From  4mVx  +  3  subtract  2aVx— 1. 

5.  From  2ax4  +  3fcc3  —  7  subtract  3mxA  —  nx3  +  2. 

6.  From  amx3  +  bnx2  -f-  ex  subtract  bmx3  —  anx2  +  ax. 

7.  From  m  + am -\-brn  subtract  am-i-bm-{-cm. 

8.  From  !  +  3acc2+5a2z3+7a3x*  subtract  x2— 3az3— 5a2x4. 


28  ALGEBRA. 

49.  Use  of  the  Parenthesis. — If  we  wish  to  indicate  that  one 
polynomial  is  to  be  subtracted  from  another,  we  may  inclose  it 
in  a  parenthesis,  and  prefix  the  sign  minus.     Thus  the  expres 
sion  7     ,  . 

a — o — (m — n-\-x) 

indicates  that  the  polynomial  m — n-\-x  is  to  be  subtracted  from 
the  polynomial  a— b.     Performing  the  operation  indicated,  we 

have  a-b-m  +  n-x. 

The  expression         a— b+(m— n+x) 

indicates  that  the  polynomial  m—n-\-x  is  to  be  added  to  the 
polynomial  a— 5,  and  the  result  is 

a— b+m— n-\-x. 

Hence  we  see  that  a  parenthesis  preceded  by  the  plus  sign 
may  be  removed  without  changing  the  signs  of  the  inclosed 
terms  ;  and,  conversely,  any  number  of  terms,  with  their  prop 
er  signs,  may  be  inclosed  in  a  parenthesis,  and  the  plus  sign 
written  before  the  whole. 

But  if  the  parenthesis  is  preceded  by  the  minus  sign,  the 
signs  of  all  the  inclosed  terms  must  be  changed  when  the  pa 
renthesis  is  removed ;  and,  conversely,  any  number  of  terms 
may  be  inclosed  in  a  parenthesis,  and  preceded  by  the  minus 
sign,  provided  the  signs  of  all  the  inclosed  terms  are  changed. 

50.  According  to  the  preceding  principle,  polynomials  may 
be  written  in  a  variety  of  forms. 

Thus,  a— b— c  +  d 

is  equivalent  to  a— (b-\-c— d}, 

or  to  a—b—(c—d), 

or  to  a  +  d—  (b  +  c). 

These  expressions  are  all  equivalent,  the  first  form  being  the 

simplest. 

EXAMPLES. 

Eeduce  the  following  expressions  to  their  simplest  forms. 

a&2). 

Ans.  a3-5 


SUBTRACTION.  29 

2.  a+b+c  —  (a—  b)—  (b—  c).  Ans.  b+2c. 

3.  4a2—  b—  (2a— 

4.  a  +  2&-3m- 

5.  3a3- 
6. 

7. 


51.  Hence  we  see  that  when  an  expression  is  inclosed  in  a 
parenthesis,  the  essential  sign  of  a  term  depends  not  merely 
upon  the  sign  which  immediately  precedes  it,  but  also  upon 
the  sign  preceding  the  parenthesis. 

Thus  m+(+n)  is  equivalent  to  m  +  n, 
and         ra~+(—  n)  m—n. 

But         m—('\-n)  "  m—  n, 

and         m—(—n)  "  m-f-rz. 

The  sign  immediately  preceding  n  is  called  the  sign  of  the 
quantity  ;  the  sign  preceding  the  parenthesis  may  be  called  the 
sign  of  the  operation  ;  while  the  sign  resulting  from  the  opera 
tion  is  called  the  essential  sign  of  the  term.  We  perceive  that 
when  the  sign  of  the  quantity  is  the  same  as  the  sign  of  opera 
tion,  the  essential  sign  of  the  term  is  positive  ;  but  when  the 
sign  of  the  quantity  is  different  from  the  sign  of  operation,  the 
essential  sign  of  the  term  is  negative. 

52.  Use  of  Negative  Quantities.  —  The  introduction  of  nega 
tive  quantities  into  Algebra  enables  us  not  only  to  compare 
the  magnitude,  but  also  to  indicate  the  relation  or  quality  of  the 
objects  about  which  we  are  reasoning.     This  peculiarity  will 
"be  understood  from  a  few  examples  : 

1st.  Gain  and  Loss  in  Trade.  —  Suppose  a  merchant  to  gain 
in  one  year  a  certain  sum,  and  in  the  following  year  to  lose  a 
certain  sum  ;  we  are  required  to  determine  what  change  has 
taken  place  in  his  capital.  This  may  be  indicated  algebraical 
ly  by  regarding  the  gains  as  positive  quantities,  and  the  losses 
as  negative  quantities.  Thus,  suppose  a  merchant,  with  a  cap 
ital  of  10,000  dollars,  loses  3000  dollars,  afterward  gains  1000 


30  ALGEBRA. 

dollars,  and  then  loses  again  4000  dollars,  the  whole  may  be 
expressed  algebraically  thus, 

10,000-3000  +  1000-4000, 

which  reduces  to  +4000.  The  +  sign  of  the  result  indicates 
that  he  has  now  4000  dollars  remaining  in  his  possession. 
Suppose  he  further  gains  500  dollars,  and  then  loses  7000  dol 
lars.  The  whole  may  now  be  expressed  thus, 

10,000-3000  +  1000-4000  +  500-7000, 

which  reduces  to  —2500.  The  —  sign  of  the  result  indicates 
that  his  losses  exceed  the  sum  of  all  his  gains  and  the  property 
originally  in  his  possession ;  that  is,  he  owes  2500  dollars  more 
than  he  can  pay. 

53.  2c?.  Motion  in  Contrary  Directions.  —  Suppose  a  ship  to 
sail  alternately  northward  and  southward,  and  we  are  required 
to  determine  the  last  position  of  the  ship.  This  may  be  indi 
cated  algebraically,  if  we  agree  to  consider  motion  in  one  direc 
tion  as  a  positive  quantity,  and  motion  in  the  opposite  direction' 
as  a  negative  quantity. 

Suppose  a  ship,  setting  out  from  the  equator,  sails  north 
ward  50  miles,  then  southward  30  miles,  then  northward  10 
miles,  then  southward  again  20  miles,  and  we  wish  to  determ 
ine  the  last  position  of  the  ship.  If  we  call  the  northerly  mo 
tion  +  ,  the  whole  may  be  expressed  algebraically  thus, 

50-30  +  10-20, 

which  reduces  to  + 10.  The  positive  sign  of  the  result  indi 
cates  that  the  ship  was  north  of  the  equator  by  10  miles. 

Suppose  the  same  ship  sails  again  40  miles  north,  then  70 
miles  south,  the  whole  may  be  expressed  thus, 

50-30  +  10-20+40-70, 

which  reduces  to  —20.     The  negative  sign  of  the  result  indi 
cates  that  the  ship  was  now  south  of  the  equator  by  20  miles. 
We  have  here  regarded  the  northerly  motion  as  + ,  and  the 


SUBTRACTION.  31 

southerly  motion  as  — ;  but  we  might  with  equal  propriety 
have  regarded  the  northerly  motion  as  — ,  and  the  southerly 
motion  as  +.  It  is,  however,  indispensable  that  we  adhere  to 
the  same  system  throughout,  and  retain  the  proper  sign  of  the 
result,  since  this  sign  shows  whether  the  ship  was  at  any  time 
north  or  south  of  the  equator. 

In  the  same  manner,  if  we  regard  westerly  motion  as  -f- ,  we 
must  regard  easterly  motion  as  — ,  and  vice  versa;  and,  gen 
erally,  when  quantities  which  are  estimated  in  different  direc 
tions  enter  into  the  same  algebraic  expression,  those  which  are 
measured  in  one  direction  being  treated  as  -f ,  those  which  are 
measured  in  the  opposite  direction  must  be  regarded  as  — . 

54.  The^same  principle  is  applicable  to  a  great  variety  of 
examples  in  Geography,  Astronomy,  etc.  Thus,  north  latitude 
is  generally  indicated  by  the  sign  +,  and  south  latitude  by  the 
sign  — .  West  longitude  is  indicated  by  the  sign  +,  and  east 
longitude  by  the  sign  — . 

Degrees  of  a  thermometer  above  zero  are  indicated  by  the 
sign  -f ,  while  degrees  below  zero  are  indicated  by  the  sign  — . 

A  variation  of  the  magnetic  needle  to  the  west  is  indicated 
by  the  sign  +,  while  a  variation  to  the  east  is  indicated  by  the 
sign  — . 

The  date  of  an  event  since  the  birth  of  Christ  is  indicated 
by  the  sign  -f  ;  the  date  of  an  event  before  the  birth  of  Christ, 
by  the  sign  —  ;  and  the  same  distinction  is  observed  in  a  great 
variety  of  cases  which  occur  in  the  application  of  the  mathe 
matics  to  practical  problems.  In  all  such  cases  the  positive 
and  negative  signs  enable  us  not  merely  to  compare  the  mag 
nitude,  but  also  to  indicate  the  relation  of  the  quantities  con 
sidered. 


32  ALGEBRA. 


CHAPTER  IV. 

MULTIPLICATION. 

55.  Multiplication  is  the  operation  of  repeating  one  quantity 
as  many  times  as  there  are  units  in  another. 

The  quantity  to  be  multiplied  is  called  the  multiplicand; 
and  that  by  which  it  is  to  be  multiplied  is  called  the  multiplier. 

56.  When  several  quantities  are  to  be  multiplied  together, 
the  result  will  be  the  same  in  whatever  order  the  multiplica 
tion  is  performed. 

In  order  to  demonstrate  this  principle,  let  unity  be  repeated 
five  times  upon  a  horizontal  line,  and  let  there  be  formed  four 
such  parallel  lines,  thus, 


Then  it  is  plain  that  the  number  of  units  in  the  table  is 
equal  to  the  five  units  of  the  horizontal  line  repeated  as  many 
times  as  there  are  units  in  a  vertical  column ;  that  is,  to  the 
product  of  5  by  4.  But  this  sum  is  also  equal  to  the  four  units 
of  a  vertical  line  repeated  as  many  times  as  there  are  units  in 
a  horizontal  line ;  that  is,  to  the  product  of  4  by  5.  Therefore 
the  product  of  5  by  4  is  equal  to  the  product  of  4  by  5.  For 
the  same  reason,  2x3x4  is  equal  to  2  x  4  x  3,  or  4  x  3  X  2,  or 
3x4x2,  the  product  in  each  case  being  24.  So,  also,  if  o,  6, 
and  c  represent  any  three  numbers,  we  shall  have  ale  equal  to 
lea  or  cab. 

CASE   I. 

When  loth  the  factors  are  monomials. 

57.  Suppose  it  is  required  to  multiply  5a  by  46.     The  prod 
uct  may  be  indicated  thus,  5a  x  4:1. 


MULTIPLICATION.  33 

But  since  the  order  of  the  factors  may  be  changed  without 
affecting  the  value  of  the  product,  the  factors  of  the  same  kind 
may  be  written  together  thus, 

4  x  Sab  ; 
or,  simplifying  the  expression,  we  have 


Hence  we  see  that  the  coefficient  of  the  product  is  equal  to  the 
product  of  the  coefficients  of  the  multiplicand  and  multiplier. 

58.  The  Law  of  Exponents. — We  have  seen,  in  Art.  16,  that 
when  the  same  letter  appears  several  times  as  a  factor  in  a 
product,  this  is  briefly  expressed  by  means  of  an  exponent. 
Thus,  aaa  is  written  a3,  the  number  3  showing  that  a  enters 
three  times  "as  a  factor.    Hence,  if  the  same  letters  are  found  in 
two  monomials  which  are  to  be  multiplied  together,  the  ex 
pression  for  the  product  may  be  abbreviated  by  adding  the 
exponents  of  the  same  letter.     Thus,  if  we  are  to  multiply  a3 
by  a2,  we  find  a3  equivalent  to  aaa,  and  a2  to  aa.     Therefore 
the  product  will  be  aaaaa,  which  may  be  written  a5,  a  result 
which  is  obtained  by  adding  together  3  and  2,  the  exponents 
of  the  common  letter  a.     Hence  we  see  that  the  exponent  of 
any  letter  in  the  product  is  equal  to  the  sum  of  the  exponents  of 
tiiis  letter  in  the  multiplicand  and  multiplier. 

59.  Hence,  for  the  multiplication  of  monomials,  we  have  the 
following 

RULE. 

Multiply  together  the  coefficients  of  the  two  terms  for  the  coeffi 
cient  of  the  product. 

Write  after  this  all  the  letters  in  the  two  monomials,  giving  to 
lack  letter  an  exponent  equal  to  the  sum  of  its  exponents  in  the  twc 
factors. 

EXAMPLES. 

1.  2.  3.  4. 

Multiply     lobe  5a2b3c  $amx*y  6a263c* 

by  5mn  8a&2c  ^am^xy3  8a3bc2 

Product    35abcp°<n         15a365c2  36a2ra3x5?/4  48a5&4c° 

B2 


34  ALGEBRA. 

5.  Multiply  9a3x  by  7a3y. 

6.  Multiply  12aW  by  llaW. 

7.  Multiply  am  by  an.  Ans.  am+n. 

8.  Multiply  8a™  by  9an. 

9.  Multiply  am  by  am.  .Arcs.  a2m. 

10.  Multiply  9am  by  12am. 

11.  Multiply  a™&  by  abm.  Ans.  am+15m+1. 

12.  Multiply  6amxn  by  5amcc.  J.ws.  30a2ma:7l+1. 

13.  Multiply  3a2mxn  by  4amx3n.  Ans.  12a3mx4ra. 

14.  What  is  the  continued  product  of  5a,  4m2#,  and  9a2m3x? 

Ans.  lSOa3m5x2. 

15.  What  is  the  continued  product  of  7a2&,  a&5,  and  4ac3F 

16.  What  is  the  continued  product  of  3amx,  5ab2,  and  lalxf 

17.  What  is  the  continued  product  of  a,  a£>,  a5c,  abed,  and 


18.  What  is  the  continued  product  of  a3,  a362,  a36c,  and 


CASE   II. 

o??e  or  loth  of  the  factors  are  polynomials. 
60.  Represent  the  sum  of  the  positive  terms  of  any  polyno 
mial  by  a,  and  the  sum  of  the  negative  terms  by  b.  Then 
a—  b  will  represent  any  polynomial  whatever.  In  like  man 
ner,  c—d  will  represent  any  other  polynomial  whatever.  It  is 
required  to  find  the  product  of  a  —  b  by  c  —  d. 

In  the  first  place,  let  us  multiply  a  —  b  by  c.  This  implies 
that  the  difference  of  the  units  in  a  and  b  is  to  be  repeated  c 
times.  If  +  a  be  repeated  as  many  times  as  there  are  units  in 
c,  the  result  will  be  -\-ac.  Also,  if  —  b  be  repeated  as  many 
times  as  there  are  units  in  c,  the  result  will  be  —  Jc,  for  —b 
taken  twice  is  —  2&,  taken  three  times  is  —  36,  etc.;  and  if  it  be 
repeated  c  times,  the  result  will  be  —  cb  or  —  be.  The  entire 
operation  may  be  exhibited  thus  : 

a-b 
c 

ac—bc. 
Next  let  us  multiply  a—  b  by  c—d.     When  we  multiply 


MULTIPLICATION.  35 

a_5  by  Cj  we  obtain  ac—bc.  But  a  —  b  was  only  to  De  taken 
c—  d  times;  therefore,  in  this  first  operation,  we  have  repeated 
it  d  times  more  than  was  required.  Hence,  to  have  the  true 
product,  we  must  subtract  d  times  a—  b  from  ac—bc.  But  c? 
times  a—  b  is  equal  to  ad—bd,  which,  subtracted  from  ac—bc, 

gives  ac-bc-ad+bd. 

If  the  pupil  does  not  perceive  the  force  of  this  reasoning,  it 
will  be  best  to  repeat  the  argument  with  numbers,  thus  :  Let  it 
be  proposed  to  multiply  8—5  by  6—2;  that  is,  the  quantity 
8—5  is  to  be  repeated  as  many  times  as  there  are  units  in 
6  —  2.  If  we  multiply  8—5  by  6,  we  obtain  48  —  30;  that  is, 
we  have  repeated  8—5  six  times.  But  it  was  only  required  to 
repeat  the  multiplicand  four  times,  or  6—2.  We  must  there 
fore  diminish  this  product  by  twice  8  —  5,  which  is  16—10; 
and  this  subtraction  is  performed  by  changing  the  signs  of  the 
subtrahend.  Hence  we  have 

48-30-16  +  10, 

which  is  equal  to  12.  This  result  is  obviously  correct;  for 
8—  5  is  equal  to  3,  and  6  —  2  is  equal  to  4;  that  is,  it  was  re 
quired  to  multiply  3  by  4,  the  result  of  which  is  12,  as  found 
above. 

We  have  thus  obtained  the  following  results  : 


•fax  (+&)= 

+  ax(—  b)=  —  ab, 
-ax(+b}=-ab, 
—  ax(—  &)= 


from  which  we  perceive  that  when  the  two  factors  have  like 
signs,  the  product  is  positive;  and  when  the  two  factors  have  un 
like  signs,  the  product  is  negative. 

61.  Hence,  for  the  multiplication  of  polynomials,  we  have 
the  following  general 

BULB. 

Multiply  each  term  of  -the  multiplicand  by  each  term  of  the  mul- 


36  ALGEBRA. 

tiplier,  and  add  together  all  the  partial  products,  observing  that 
like  signs  require  +  in  the  product,  and  unlike  signs  — . 

EXAMPLES. 
1.  2. 

Multiply     2a  +  Sb  a2-f-2a&-f&2 

by  4a— 56  a+b 


Partial    (8a2+12a&  a3+2a2£  +  ab2 

products   (_    -Wab-15b* 
Eesult         8a2  +   2a&-15ia 


It  is  immaterial  in  what  order  the  terms  of  a  polynomial 
are  arranged,  or  in  what  order  the  letters  of  a  term  are  ar 
ranged.  It  is,  however,  generally  most  convenient  to  arrange 
the  letters  of  a  term  alphabetically,  and  to  arrange  the  terms 
of  a  polynomial  in  the  order  of  the  powers  of  some  common 
letter. 

3.  Multiply  a2—  ab  +  b2  by  a-\-b.  Ans.  a3-f  &3. 

4.  Multiply  a2-2ab+b2  by  a  -b. 

5.  Multiply  3a2-2a+5  by  a—  4. 

6.  Multiply  cP  —  db  +  W  by  a2  +  ab  +  b2.     Ans.  a4  +  a2b2+b*. 

7.  Multiply  2a2—  3a&  +  4  by  a2-f  2a&—  3. 

8.  Multiply  a3  +  a26  +  a&2  +  &3  by  a—b. 

9.  Multiply  a  +  mb  by  a  +  ??5. 

10.  Multiply  3a+2fcc-3x2  by  3a-2bx+3x*. 

11.  Multiply  together  £—5,  x-f  2,  and  x+3. 

12.  Multiply  together  x  —  3,  x  —  4,  x-j-5,  and  cc  —  6. 

13.  Multiply  together  a2+a6  +  Z>2,  a2—  ab  +  b2,  and  a2-62. 

14.  Multiply  together  a  +  x,  b-\-x,  and  c-\-x. 

15.  Multiply  a4  +  a3&  +  a2£2  +  «&3  +  64  by  a-b. 

16.  Multiply  a3-3a2  +  7a-12  by  a2+3a+2. 

17.  Multiply  x4  +  2x3  +  3x2  +  2x+l  by  x2—  2x+l. 

18.  Multiply  14a3x—  6a2bx+x2  by  14a3x+6a2^—  a;2. 

19.  Multiply  x3—x2y+xy2  by  x2—xif—y*. 

20.  Multiply  3x2+8x?/-5  by  4x2-7xi/4-  9. 

62.  Degree  of  a  Product.  —  Since,  in  the  multiplication  of  two 
monomials,  every  factor  of  both  quantities  appears  in  the  prod 


MULTIPLICATION.  37 

uct,  it  is  obvious  that  the  degree  of  the  product  will  be  equal 
to  the  sum  of  the  degrees  of  the  multiplier  and  multiplicand. 
Hence,  also,  if  two  polynomials  are  homogeneous,  their  product 
will  be  homogeneous. 

Thus,  in  the  first  example  of  the  preceding  article,  each 
term  of  the  multiplicand  is  of  the  first  degree,  and  also  each 
term  of  the  multiplier;  hence  each  term  of  the  product  is  of 
the  second  degree.  For  a  similar  reason,  in  the  second  exam 
ple,  each  term  of  the  product  is  of  the  third  degree ;  and  in 
the  sixth  example,  each  term  of  the  product  is  of  the  fourth 
degree.  This  principle  will  assist  us  in  guarding  against  er 
rors  in  the  multiplication  of  polynomials,  so  far  as  concerns 
the  exponents. 

63.  Number  of  Terms  in  a  Product. — When  the  product  aris 
ing  from  the  multiplication  of  two  polynomials  does  not  admit 
of  any  reduction  of  similar  terms,  the  whole  number  of  terms 
in  the  product  is  equal  to  the  product  of  the  numbers  of  the 
terms  in  the  two  polynomials.     Thus,  if  we  have  five  terms  in 
the  multiplicand,  and  four  terms  in  the  multiplier,  the  whole 
number  of  terms  in  the  product  will  be  5x4,  or  20.     In  gen 
eral,  if  there  be  m  terms  in  the  multiplicand,  and  n  terms  in 
the  multiplier,  the  whole  number  of  terms  in  the  product  will 
be  mxn. 

64.  Least  Number  of  Terms  in  a  Product. — If  the  product  of 
two  polynomials  contains  similar  terms,  the  number  of  terms 
in  the  product,  when  reduced,  may  be  much  less  than  mn ; 
but  it  is  important  to  observe  that  among  the  different  terms 
of  the  product  there  are  always  two  which  can  not  be  combined 
with  any  others.     These  are, 

1st.  The  term  arising  from  the  multiplication  of  the  two 
terms  affected  with  the  highest  exponent  of  the  same  letter. 

2d.  The  term  arising  from  the  multiplication  of  the  two 
terms  affected  with  the  lowest  exponent  of  the  same  letter. 

For  it  is  evident,  from  the  rule  of  exponents,  that  these  iwo 
partial  products  must  involve  the  letter  in  question,  the  one 


38  ALGEBRA. 

with  a  higher,  and  the  other  with  a  lower  exponent  than  any 
of  the  other  partial  products,  and  therefore  can  riot  be  similar 
to  any  of  them.  Hence  the  product  of  two  polynomials  can  never 
contain  less  than  tivo  terms. 

65.  For  many  purposes  it  is  sufficient  merely  to  indicate  the 
multiplication  of  two  polynomials,  without  actually  perform 
ing  the  multiplication.  This  is  effected  by  inclosing  the  poly 
nomials  in  parentheses,  and  writing  them  in  succession,  with 
or  without  the  sign  x  .  When  the  indicated  multiplication 
has  been  actually  performed,  the  expression  is  said  to  be  ex 
panded. 

EXAMPLES. 

1.  Expand  (a+b)  (c-{-d).  Ans.  ac-\-bc  +  a 

2.  Expand  9a-7(6-c). 

3.  Expand  and  reduce 


4.  Expand  and  reduce 

28(a-£  +  c)+ 

5.  Expand  and  reduce 

24a-6&-9(a+&) 

6.  Expand  and  reduce 


66.  The  three  following  theorems  have  very  important  ap 
plications. 

The  square  of  the  sum  of  two  numbers  is  equal  to  the  square  of 
the  first,  plus  twice  the  product  of  the  first  by  the  second,  plus  the 
square  of  the  second. 

Thus,  if  we  multiply      a-\-b 
by  a  +  b 

a'z  --   ab 


we  obtain  the  product        a2 

Hence,  if  we  wish  to  obtain  the  square  of  a  binomial,  we 
can,  according  to  this  theorem,  write  out  at  once  the  term,? 


MULTIPLICATION.  39 

of  the  result,  without  the  necessity  of  performing  an  actual 
multiplication. 

EXAMPLES. 

1.  (3a  +  Z>)2^  6.  (5a24-7aZ>)2=: 

3.  (5a-J-3&)2-  8.  (2a  +  i)2- 

4.  (5a2-f2£)2  =  9.  (l+^)2  = 

5.  (5a3+b}2  =  10. 


67.  The  square  of  the  difference  of  two  numbers  is  equal  to  ih& 
square  of  the  first,  minus  twice  the  product  of  the  first  by  the  sec 
ond,  plus  the  square  of  the  second. 

Thus,  i£  we  multiply      a — b 
by  a— b 

a2 —   ab 


we  obtain  the  product        a2—2ab-\-b2. 

EXAMPLES. 

1.  (2a— 3b)2  =  6.  (7a2— 12«6)2  — 

2.  (5a— ±b)2=  7.  (7a262—  12a^)2  = 

3.  (6«2_x)2=  8.  (2a3-5)2=r 

4.  (6a2— Sx)2=  9.  (2—£)2= 

68.  Meaning  of  the  sign  ±. 

Since  (a  +  b)2  =  a'2-}-2ab-\-b2, 

and  (a—b)2  =  a2—2ab-\-b'2J 

we  may  write  both  formulas  in  the  following  abbreviated  form, 


which  indicates  that,  if  we  use  the  +  sign  of  b  in  the  root,  we 
must  use  the  +  sign  of  2ab  in  the  square ;  but  if  we  use  the 
—  sign  of  b  in  the  root,  we  must  use  the  —  sign  of  2ab  in  the 
square.  By  this  notation  we  are  enabled  to  express  two  dis 
tinct  theorems  by  one  formula. 


40  ALGEBRA. 

69.  The  product  of  the  sum  and  difference  of  two  numbers  is 
equal  to  the  difference,  of  their  squares. 

Thus,  if  we  multiply       a-j-5 
by  a  —  b 

a?+ab 

-ab-b* 
tfe  obtain  the  product        a2        —  b2. 

EXAMPLES. 

1.  (3a+26)  (3a-26)= 

2.  (lab+x)  (7ab-x)  = 

3.  (8a+7bc)  (8a-7bc)  = 
4. 

5. 

6. 

7. 

8.  (4+|)  (4-*)= 

The  pupil  should  be  drilled  upon  examples  like  the  pre 
ceding  until  he  can  produce  the  results  mentally  with  as  great 
facility  as  he  could  read  them  if  exhibited  upon  paper,  and 
without  committing  the  common  mistake  of  making  the  square 
of  a +  6  equal  to  a2  +  62,  or  the  square  of  a  —  b  equal  to  a?—b2. 
The  utility  of  these  theorems  will  be  the  more  apparent 
when  they  are  applied  to  very  complicated  expressions.  Fre 
quent  examples  of  their  application  will  be  seen  hereafter. 


DIVISION.  41 


CHAPTER  Y. 

DIVISION. 

70.  Division  is  the  converse  of  multiplication.     In  multipli 
cation  we  determine  the  product  arising  from  two  given  fac 
tors.     In  division  we  have  the  product  and  one  of  the  factors 
given,  and  we  are  required  to  determine  the  other  factor. 

The  dividend  is  the  product  of  the  divisor  and  quotient,  the 
divisor  is  the  given  factor,  and  the  quotient  is  the  factor  re 
quired  to  be  found. 

When  the  divisor  and  dividend  are  loth  monomials. 

71.  Since  the  product  of  the  numbers  denoted  by  a  and  b  is 
denoted  by  a&,  the  quotient  of  ab  divided  by  a  is  b;  that  is, 
ab-^-a^b.    Similarly,  we  have  abc^-a=bc,  abc^rc=ab,  abc-^ab 
=c,  etc.     The  division  is  more  commonly  denoted  thus: 

abc     .          abc  _  abc  _   , 

a  ~  b  c 

abc  abc     7          abc 

So,  also,  12ran  divided  by  3m  gives  &i;  for  3m  multiplied 
by  4:n  makes  12m?2. 

72.  Rule  of  Exponents  in  Division. — Suppose  we  have  a5  to 
be  divided  by  a2.     We  must  find  a  quantity  which,  multiplied 
by  a2,  will  produce  a5.     We  perceive  that  a3  is  such  a  quanti 
ty  ;  for,  according  to  Art.  58,  in  order  to  multiply  a3  by  a2,  we 
add  the  exponents  2  and  3,  making  5 ;  that  is,  the  exponent  3 
of  the  quotient  is  found  by  subtracting  2,  the  exponent  of  the 
divisor,  from  5,  the  exponent  of  the  dividend. 

Hence,  in  order  to  divide  one  power  of  any  quantity  by  an 
other  power  of  the  same  quantity,  subtract  the  exponent  of  the 
divisor  from  the  exponent  of  the  dividend. 


42  ALGEBRA. 

73.  Proper  Sign  of  the  Quotient. — The  proper  sign  to  be  pre 
fixed  to  a  quotient  may  be  deduced  from  the  principles  al 
ready  established  for  multiplication.  The  product  of  the  di 
visor  and  quotient  must  be  equal  to  the  dividend.  Hence, 

because  +  ax(+&)=+a 

—  ax(  +  b)=—ab.       ,       c         } 

;     ,(  7'     therefore 

+  ax(— b)=—  ab, 


+ab^—b=— 


-ax(-b)=+ab, 

Hence,  if  the  dividend  and  divisor  have  like  signs,  the  quotient 
will  be  positive  ;  but  if  they  have  unlike  signs,  the  quotient  will  be 
negative. 

74.  Hence,  for  dividing  one  monomial  by  another,  we  have 
the  following 

RULE. 

1.  Divide  the  coefficient  of  the  dividend  by  the  coefficient  of  the 
divisor,  for  a  new  coefficient. 

2.  To  this  result  annex  all  the  letters  of  the  dividend,  giving  to 
each  an  exponent  equal  to  the  excess  of  its  exponent  in  the  dividend 
above  that  in  the  divisor. 

3.  If  the  dividend  and  divisor  have  like  signs,  prefix  the  plus 
sign  to  the  quotient;   but  if  they  have  unlike  signs,  prefix  the 
minus  sign. 

EXAMPLES. 

1.  Divide  20ax3  by  4o?.  Arts.  Sax"2. 

2.  Divide  25&3x?/4  by  —  6ay2.  Ans.  —  5a2xy2. 

3.  Divide  —  72ab5x2  by  I2b3x.  Ans.  —  6ab2x. 

4.  Divide  -77a3//c6  by  -Uab3c\  Ans. 

5.  Divide  48aWd  by  -I2ab2c. 

6.  Divide  -I50a5b9cd3  by  30aW2. 

7.  Divide  -250a7£V  by  -5abx3. 

8.  Divide  272aWx6  by  -I7a2b3cx*. 

9.  Divide  -42a663c  by  21a63a 
10.  Divide  -SOQa3b*x  by  -50bx. 


DIVISION.  43 

75.  Value  of  the  Symbol  a°. — The  rule  given  in  Art.  72  con 
ducts  us  in  some  cases  to  an  expression  of  the  form  a°.     Let 
it  be  required  to  divide  a2  by  a2.     According  to  the  rule,  the 
quotient  will  be  a2"2,  or  a°.     Now  every  number  is  contained 
in  itself  once ;  hence  the  value  of  the  quotient  must  be  unity ; 
that  is,  a°  =  l. 

To  demonstrate  this  principle  generally,  let  a  represent  any 
quantity,  and  m  the  exponent  of  any  power  whatever.  Then, 
by  the  rule  of  division, 

a™  •+•  a™ =a'm-m=a(). 

But  the  quotient  obtained  by  dividing  any  quantity  by  it 
self  is  unity ;  that  is,  a°  =  l, 
or  any  quantity  having  a  cipher  for  its  exponent  is  equal  to  unity. 

76.  Signification  of  Negative  Exponents. — The  rule  given  in 
Art.  72  conducts   us  in   some   cases  to   negative   exponents. 
Thus,  let  it  be  required  to  divide  a3  by  a5.    We  are  directed 
to  subtract  the  exponent  of  the  divisor  from  the  exponent  of 
the  dividend.     We  thus  obtain 

a3"5,  or  a~2. 

a3 

But  a3  divided  by  a5  may  be  written  -gj    and,  since  the 

ct 

value  of  a  fraction  is  not  altered  by  dividing  both  numerator 
and   denominator  by  the   same   quantity,  this  expression  is 

equivalent  to  — . 

Hence  a~2  is  equivalent  to  — =. 

az 

So,  also,  if  a2  is  to  be  divided  by  a5,  this  may  be  written 
a2     1 


~~A —  — 5  —  a 
5       3 


-3 


a"     aj 
In  the  same  manner,  we  find 

—  =a-m- 

that  is,  any  quantity  having  a  negative  exponent  is  equal  to  the 
reciprocal  of  that  quantity  with  an  equal  positive  exponent. 


44  ALGEBKA. 

77.  Hence  any  factor  may  be  transferred  from  the  numera 
tor  to  the  denominator  of  a  fraction,  or  from  the  denominatoi: 
to  the  numerator,  by  changing  the  sign  of  its  exponent. 

Thus,  ^  may  be  written  ab~\ 
g  "  a^», 


that  is,  the  denominator  of  a  fraction  may  be  entirely  re 
moved,  and  an  integral  form  be  given  to  any  fractional  ex 
pression. 

This  use  of  negative  exponents  must  be  understood  simply 
as  a  convenient  notation,  and  not  as  a  method  of  actually  de 
stroying  the  denominator  of  a  fraction. 

78.  To  divide  a  Polynomial  by  a  Monomial.  —  We  have  seen, 
Art.  60,  that  when  a  single  term  is  multiplied  into  a  polyno 
mial,  the  former  enters  into  every  term  of  the  latter. 

Thus,  (a+b)m=am+bm; 

therefore  (am-\-bm)~m=a-\-b. 

Hence,  to  divide  a  polynomial  by  a  monomial,  we  have  the 
following 

RULE. 

Divide  each  term  of  the  dividend  by  the  divisor,  ^nd  connect  the 
quotients  by  their  proper  signs. 

EXAMPLES. 

1.  Divide  3cc3  +  6x2+3a:e—  I5x  by  3x.     Ans.  a;2  4-  2^+^—  5. 
Z  Divide  3abc+I2abx—  9a2b  by  Sab.  Ans,  c-f  4cc  —  3o- 

3.  Divide  40a3&3+60a262-17a&  by  -ab. 

4.  Divide  15a2bc—  Wacx2  +  Sa&d2  by  —  5a2c. 

5.  Divide  20;r5-35x4-15x3+75a;2  by  -5x2. 

6.  Divide  6a2x4z/6-12a3xy  +  15a4cc5?/3  by  3a2x2?/2. 

7.  Divide  xn+l  —  xn+*  +  xn+3  —  xn+4  by  xn. 

8.  Divide  12ay-16a5/  +  20ay-28ay  by  -4ay. 


DIVISION.  45 

79.  To  divide  one  polynomial  ~by  another. 
Let  it  be  required  to  divide 

2a6-f  a2+62  by  a+b. 

The  object  of  this  operation  is  to  find  a  third  polynomial 
which,  multiplied  by  the  second,  will  reproduce  the  first. 

It  is  evident  that  the  dividend  is  composed  of  all  the  partial 
products  arising  from  the  multiplication  of  each  term  of  the 
divisor  by  each  term  of  the  quotient,  these  products  being 
added  together  and  reduced.  Hence,  if  we  can  discover  a 
term  of  the  dividend  which  is  derived  without  reduction  from 
the  multiplication  of  a  term  of  the  divisor  by  a  term  of  the 
quotient,  then  dividing  this  term  by  the  corresponding  term 
of  the  divisor,  we  shall  be  sure  to  obtain  a  term  of  the  quo 
tient. 

But,  from  Art.  64,  it  appears  that  the  term  a2,  which  con 
tains  the  highest  exponent  of  the  letter  a,  is  derived  without  re 
duction  from  the  multiplication  of  the  two  terms  of  the  divisor 
and  quotient  which  are  affected  with  the  highest  exponent  of 
the  same  letter.  Dividing  the  term  a2  by  the  term  a  of  the 
divisor,  we  obtain  a,  which  we  are  sure  must  be  one  term  of 
the  quotient  sought.  Multiplying  each  term  of  the  divisor  by 
a,  and  subtracting  this  product  from  the  proposed  dividend, 
the  remainder  may  be  regarded  as  the  product  of  the  divisor 
by  the  remaining  terms  of  the  quotient.  We  shall  then  ob 
tain  another  term  of  the  quotient  by  dividing  that  term  of  the 
remainder  which  is  affected  with  the  highest  exponent  of  a  by 
the  term  a  of  the  divisor,  and  so  on. 

Thus  we  perceive  that  at  each  step  we  are  obliged  to  search 
for  that  term  of  the  dividend  which  is  affected  with  the  high 
est  exponent  of  one  of  the  letters,  and  divide  it  by  that  term  of 
the  divisor  which  is  affected  with  the  highest  exponent  of  tha 
same  letter.  We  may  avoid  the  necessity  of  searching  for  this 
term  by  arranging  the  terms  of  the  divisor  and  dividend  in 
ike  order  of  the  powers  of  one  of  the  letters. 

The  operation  will  then  proceed  as  follows: 


4:6  ALGEBRA. 

The  arranged  dividend  is  gP+Zab+b*  I  a-\-b,  the  divisor. 

ft2+  ab     _  I  a+Vthe  quotient. 
ab  +  b2,  the  first  remainder. 


0,  remainder. 

For  convenience  of  multiplication,  the  divisor  is  written  on 
the  right  of  the  dividend,  and  the  quotient  under  the  divisor, 

80.  Hence,  to  divide  one  polynomial  by  another,  we  have 
the  following 

RULE. 

1.  Arrange  loth  polynomials  in  the  order  of  the  powers  of  the 
same  letter. 

2.  Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor,  for  the  first  term  of  the  quotient. 

3.  Multiply  the  whole  divisor  by  this  term,  and  subtract  the 
product  from  the  dividend. 

4.  Divide  the  first  term  of  the  remainder  by  the  first  term  of  the 
divisor,  for  the  second  term  of  the  quotient. 

5.  Multiply  the  whole  divisor  by  this  term,  and  subtract  the 
product  from  the  last  remainder. 

6.  Continue  the  same  operation  until  a  remainder  is  found  equal 
to  zero,  or  one  whose  first  term  is  not  divisible  by  the  first  term  of 
the  divisor. 

When  a  remainder  is  found  equal  to  zero,  the  division  is 
said  to  be  exact.  When  a  remainder  is  found  whose  first 
term  is  not  divisible  by  the  first  term  of  the  divisor,  the  exact 
division  is  impossible.  In  such  a  case,  the  last  remainder  must 
be  placed  over  the  divisor  in  the  form  of  a  fraction,  and  an 
nexed  to  the  quotient. 

EXAMPLES. 

1.  Divide  2a2&  +  &3  +  2a&2  +  «3  by  a2  +  b2  +  ab.         Ans.  a  +  b. 

2.  Divide  x3  —  a3  +  3a2x—  3ax2  by  x—  a.    Ans.  x2—2ax-{-az. 
B.  Divide  a6  +  «6  +  2a3a;3  by  a2—  ax+x2. 

Ans.  «4-f 


DIVISION.  47 


4.  Divide  a6-16a3x3  +  64x6  by  a2- 

5.  Divide  a4-f-6a2x2—  4a3x+x4—  4ax3  by  a2— 

.Afts.  a2  — 

6.  Divide  32x5  +  ?/5  by  2a?+y. 

7.  Divide  x5  +  x3?/2-fx?/4—  x4?/  —  x2y3  —  y5  by  x^— 

8.  Divide  x4  +  x3-f  5x—  4x2—  3  by  a2—  2x—  d. 

9.  Divide  a6-b6  by  a3  +  2a26-f  2a 

10.  Divide  x6  +  l-2x3  by  cc2  +  l- 

11.  Divide  x*  +  y*+xzy2  by  x2  + 

12.  Divide  12x4-192  by  3x-6. 

Ans.  4x3  +  8x2  +  16x+32. 

13.  Divide  6x6-6/  by  2^c2-2?/2. 

14.  Divide  a6  +  3a264-3a462-^6  by  a3- 


15.  Divide  x6-6x4-f  9cc2-4  by  #2-l. 

16.  Divide  a4  +  a36  —  8a2^2  +  19«63—  15Z>4  by 

17.  Divide  x3  +  ?/3  +  3x?/—  1  by  x+y—l. 

18.  Divide  a2&2+2a&c2—  a2c2  —  62c2  by  ab  +  ac  —  bc. 

19.  Divide  a3-63  by  a  -b. 

20.  Divide  a4-64  by  a-b. 

81.  Hitherto  we  have  supposed  the  terms  of  the  quotient  to 
be  obtained  by  dividing  that  term  of  the  dividend  which  is 
affected  with   the   highest  exponent  of  a  certain  letter.     But, 
from  Art.  64,  it  appears  that  the  term  of  the  dividend  affected 
with  the  lowest  exponent  of  any  letter  is  derived  without  re 
duction  from  the  multiplication  of  a  term  of  the  divisor  by  a 
term  of  the  quotient.     Hence  we  may  obtain  a  term  of  the 
quotient  by  dividing  the  term  of  the  dividend  affected  with 
the  lowest  exponent  of  any  letter  by  the  term  of  the  divisor 
containing  the  lowest  exponent  of  the  same  letter;   and  we 
may  even  operate  upon  the  highest  and  lowest  exponents  of  a 
certain  letter  alternately  in  the  same  example. 

82.  an  —  bn  is  always  divisible  by  a—b.    From  the  examples 
of  Art.  80  we  perceive  that  a3  —  b3  is  divisible  by  a  —  b;  and 
ft*_/;4  is  divisible  by  a—b.     We  shall  find  the  same  to  hold 


4:8  ALGEBRA. 

true,  whatever  may  be  the  value  of  the  exponents  of  the  two 
letters ;  that  is,  the  difference  of  the  same  powers  of  any  two  quan 
tities  is  always  divisible  by  the  difference  of  the  quantities. 
Thus,  let  us  divide  a5  —  b5  by  a— b: 

a5 — b5      a — 5,  divisor. 

a4,     partial  quotient 


The  first  term  of  the  quotient  is  a4,  and  the  first  remainder 
is  a46— Z>5,  which  may  be  written 

b(a*-b*). 

Now  if,  after  a  division  has  been  partially  performed,  the 
remainder  is  divisible  by  the  divisor,  it  is  obvious  that  the 
dividend  is  completely  divisible  by  the  divisor.  But  we  have 
already  found  that  a4  —  b*  is  divisible  by  a^b  ;  therefore  a5  —  b5 
is  also  divisible  by  a — b  ;  and,  in  the  same  manner,  it  may  be 
proved  that  aG—b6  is  divisible  by  a— 6,  and  so  on. 

83.  To  exhibit  this  reasoning  in  a  more  general  form,  let  n 
represent  any  positive  whole  number  whatever,  and  let  us  at 
tempt  to  divide  an— bn  by  a— b.  The  operation  will  be  as 

follows ; 

an—bn          a  — Z>,  divisor. 

an  —  ban-1  I  a71"1,  quotient. 
The  first  remainder  is      ban~1—bn. 

Dividing  an  by  a,  we  have,  by  the  rule  of  exponents,  an~l 
for  the  quotient.  Multiplying  a— b  by  this  quantity,  and  sub 
tracting  the  product  from  the  dividend,  we  have  for  the  first 
remainder  ban~l  —  bn,  which  may  be  written 

b(an-^-bn~l). 

Now,  if  this  remainder  is  divisible  by  a  —  b,  it  is  obvious 
chat  the  dividend  is  divisible  by  a  —  b  ;  that  is,  if  the  difference 
of  the  same  powers  of  two  quantities  is  divisible  by  the  difference  of 
the  quantities,  then  will  the  difference  of  the  powers  of  the  next 
higher  degree  be  divisible  by  that  difference. 

Therefore,  since  a4 — b4  is  divisible  by  a— &,  a5  —  bs  must  be 


DIVISION.  49 

divisible  by  a—  b;   also  a6—  Z>6,  and  so  on  for  any  positive 
value  of  n. 

The  quotients  obtained  by  dividing  the  difference  of  the 
same  powers  of  two  quantities  by  the  difference  of  the  quan 
tities  follow  a  simple  law.  Thus, 


(a6  -  b5)  -r-  (a  -  b)  =  a*  +  a3b  +  a2b2  +  ab3  +  64, 
etc.  etc.  etc. 


The  exponents  of  a  decrease  by  unity,  while  those  of  b  in 
crease  by  unity. 

84.  It  may  also  be  proved  that  the  difference  of  like  even  pow 
ers  of  any  two  quantities  is  always  divisible  by  the  sum  of  the 
quantities. 

Thus,  (a2-b2)^(a  +  b)  =  a-b. 


(a6  _  b6)  -r  (a  +  b)  =  a5  -  a*b  +  a3b2  -  a2b3  +  ab*  -  b5, 
etc.  etc.  etc. 

Also,  the  sum  of  like  odd  powers  of  any  two  quantities  is  always 
divisible  by  the  sum  of  the  quantities. 
Thus,  (a3 


(a7  +  67)  H-  (a  +  6)  =  a6  -  a5b  +  a4/>2  -  a3b3  -f  a2Z>4  -  a  65  +  56, 

etc.  etc.  etc. 

The  exponents  of  a  and  b  follow  the  same  law  as  in  Art.  83, 

but  the  signs  of  the  terms  are  alternately  plus  and  minus. 

* 

85.  When  exact  division  is  impossible.  —  One  polynomial  can 
not  be  divided  by  another  polynomial  containing  a  letter  which 
is  not  found  in  the  dividend;  for  it  is  impossible  that  one 
quantity  multiplied  by  another  which  contains  a  certain  letter 
should  give  a  product  not  containing  that  letter. 

C 


50  ALGEBRA. 

A  monomial  is  never  divisible  by  a  polynomial,  because 
every  polynomial  multiplied  by  another  quantity  gives  a  prod 
uct  containing  at  least  two  terms  not  susceptible  of  reduction. 

Yet  a  binomial  may  be  divided  by  a  polynomial  containing 
kny  number  of  terms. 

Thus,  a4— &4  is  divisible  by  a?  +  a2b  +  ab2+b3,  and  gives  for 

quotient  a— b. 

To  resolve  a  Polynomial  into  Factors. 

86.  When  a  polynomial  is  capable  of  being  resolved  into 
factors,  the  factors  can  generally  be  discovered  by  inspection, 
or  from  the  law  of  formation. 

If  all  the  terms  of  a  polynomial  have  a  common  factor,  that 
factor  is  a  factor  of  the  polynomial ;  and  the  other  factor  may 
be  found  by  dividing  the  polynomial  by  the  common  factor. 

EXAMPLES. 

1.  Eesolve  3a262  +  3a63+3a£2c  into  factors. 

Ans.  3ab2(a  +  b+c). 

2.  Eesolve  5a4b2-Wa3b3-5a2b*-5a2b2  into  factors. 

Ans.  5a2b2(a2-2ab-b2-l). 

3.  Eesolve  Qa2b2c3  —  I2ab2c3mx—  I8ab2c3y  into  factors. 

Ans.  6ab2c3(a-2mx-Sy). 

4.  Eesolve  7a3b2  —  7a2b3  —  7a2b2c  into  factors. 

5.  Eesolve  8a2bc+l2ab2c— I6abc2  into  factors. 

6.  Eesolve  10ab2cmx—5ab2cy-\-5ab2c  into  factors. 

87.  When  two  terms  of  a  trinomial  are  perfect  squares,  and 
tke  third  term  is  twice  the  product  of  their  square  roots,  the 
trinomial  will  be  the  square  of  the  sum  or  difference  of  these 
roots,  Arts.  66  and  67,  and  may  be  resolved  into  factors  ac 
cordingly. 

EXAMPLES. 

1.  Eesolre  ji2— 2ab-{-b2  into  factors.      Ans.  (a  —  b)(a  —  b). 

2.  Eesolve  «2  +  4a^4-4^2  into  factors. 

Ans.  (a  +  25)  (a  +  25). 

3.  Eesolve  a2—  6ab  +  9b2  into  factors. 


DIVISION.  61 

4.  Kesolve  9a2— 24a&4-1662  into  factors. 

5.  Kesolve  25a4— 60a2&3  +  3666  into  factors. 

6.  Kesolve  kmW— 4mn-f-l  into  factors. 

7.  Kesolve  49a464— 168a363+144a262  into  factors. 

8.  Kesolve  ^3+2n2+w  into  three  factors. 

9.  Resolve  16a462— 24a2&rax+9?ft2x2  into  factors. 
10.  Kesolve  m47i2+2m3^3-|-m2ni  into  three  factors. 

88.  If  a  binomial  consists  of  two  squares  connected  by  the 
minus  sign,  it  must  be  equal  to  the  product  of  the  sum  and 
difference  of  the  square  roots  of  the  two  terms,  Art.  69,  and 
may  be  resolved  into  factors  accordingly. 

EXAMPLES. 

1.  Kesolve  4a2-952  into  factors.      Ans.  (2a+36)  (2a— 3&> 

2.  Kesolve  9a262— 16a2c2  into  factors. 

3.  Kesolve  a5x— 9ax3  into  three  factors. 

4.  Kesolve  a4— b*  into  three  factors. 

5.  Kesolve  a6—b6  into  its  factors. 

6.  Kesolve  a8— b9  into  four  factors. 

7.  Kesolve  1—  TV  into  two  factors. 

8.  Resolve  4  —  TV  into  two  factors. 

89.  If  the  two  terms  of  a  binomial  are  both  powers  of  the 
same  degree,  it  may  generally  be  resolved  into  factors  accord 
ing  to  the  principles  of  Arts.  82-84. 

EXAMPLES. 

1.  Resolve  a?  —  b3  into  its  factors.     Ans.  (a2 -\-ab-\-b2)  (a^b). 

2.  Resolve  a3  +  b3  into  its  factors. 

3.  Resolve  a6  —  b6  into  four  factors. 

4.  Resolve  a3  —  Sb3  into  its  factors, 

5.  Resolve  8a3— 1  into  its  factors. 

6.  Resolve  8a3— Sb3  into  three  factors. 

7.  Resolve  1  +  2763  into  its  factors. 

8.  Resolve  8a3  +  27&3  into  its  factors. 

9.  Resolve  a16  —  b16  into  five  factors. 


62  ALGEBRA. 


CHAPTEK  VL-r 

GEEATEST  COMMON  DIVISOR.  —  LEAST  COMMON  MULTIPLE. 

90.  A  common  divisor  of  two  quantities  is  a  quantity  which 
will  divide  them  both  without  a  remainder.  Thus  2ab  is  a 
common  divisor  of  6a?b2x  and 


91.  A  prime  factor  is  one  that  can  not  be  resolved  into  any 
other  factors.     It  is,  therefore,  divisible  only  by  itself  and  uni 
ty.     Thus  the  quantity  2a2—  2ab  is  the  product  of  the  three 
prime  factors  2,  a,  and  a—  b. 

92.  The  greatest  common  divisor  of  two  quantities  is  the 
greatest  quantity  which  will  divide  each  of  them  without  a 
remainder.     It  is  the  continued  product  of  all  the  prime  fac 
tors  which  are  common  to  both.     The  term  greatest  here  refers 
to  the  degree  of  a  quantity,  or  of  its  leading  term,  and  not  to  its 
arithmetical  value. 

93.  When  both  quantities  can  be  resolved  into  prime  fac 
tors  by  methods  already  explained,  the  greatest  common  di 
visor  may  be  found  by  the  following 

RULE. 

Resolve  both  quantities  into  their  prime  factors.  The  continued 
product  of  all  those  factors  which  are  common  to  bothj  will  be  the 
greatest  common  divisor  required. 

EXAMPLES. 

1.  Find  the  greatest  common  divisor  of  4a2fce  and  6ab2x3. 
Kesolving  into  factors,  we  have 

4a2fce  =  2a  X  2a  x  b  x  x. 
6ab2x3  =  2a  xSbxbxxXxxx. 

The  common  factors  are  2a,  6,  and  x.  Hence  the  greatest 
common  divisor  is  2abx. 


GREATEST  COMMON  DIVISOK.  53 

2.  Find  the  greatest  common  divisor  of  4am2 -f  45m2  and 
Ban  +  3bn. 

Kesolving  into  factors,  we  have 

4am2 + 45m2 = 2m  x  2m  (a + b). 

3an+3bn=3n(a+b). 
Hence  a +b  is  the  greatest  common  divisor. 

3.  Find  the  greatest  common  divisor  of  x3—^3  and  y?—y\ 

x3—y3  =  (x—y)(x2-{-xy-{-y'2). 


Hence  x— y  is  the  greatest  common  divisor. 

4.  Find   the   greatest   common   divisor  of  35a2bmx2   and 
42am2cc3. 

5.  Find  the  greatest  common  divisor  of  3a2x—6abx-4-3b2x 
and  4a2y — 43)2y. 

6.  Find  the  greatest  common  divisor  of  9mx2— 6mx  +  m 
and  9?ix2— n. 

7.  Find  the  greatest  common  divisor  of  12a2— 36a5-}-2763 
and  8a2-1862. 

94.  When  the  given  quantities  can  not  be  resolved  into 
prime  factors  by  inspection,  the  greatest  common  divisor  may 
be  found  by  applying  the  following  principle : 

The  greatest  common  divisor  of  two  quantities  is  the  same  with 
the  greatest  common  divisor  of  the  least  quantity,  and  their  remain 
der  after  division. 

To  prove  this  principle,  let  the  greatest  of  the  two  quanti 
ties  be  represented  by  A,  and  the  least  by  B.  Divide  A  by 
B;  let  the  entire  part  of  the  quotient  be  represented  by  Q, 
and  the  remainder  by  R.  Then,  since  the  dividend  must  be 
equal  to  the  product  of  the  divisor  by  the  quotient,  plus  the 
remainder,  we  shall  have  A—  QB  +  R. 

Now  every  number  which  will  divide  B  will  divide  QB; 
and  every  number  which  will  divide  R  and  QB  will  divide 
R+QB,  or  A.  That  is,  every  number  which  is  a  common 
divisor  of  B  and  R  is  a  common  divisor  of  A  and  B. 

Again :  every  number  which  will  divide  A  and  B  will  di- 


54  ALGEBRA. 

vide  A  and  QB ;  it  will  also  divide  J.—  QB,  or  E.  That  is, 
every  number  which  is  a  common  divisor  of  A  and  B  is  also 
a  common  divisor  of  B  and  R.  Hence  the  greatest  'common 
divisor  of  A  and  B  must  be  the  same  as  the  greatest  common 
divisor  of  B  and  R. 

95.  To  find,  then,  the  greatest  common  divisor  of  two  quanti 
ties,  we  divide  the  greater  by  the  less;  and  the  remainder, 
which  is  necessarily  less  than  either  of  the  given  quantities, 
is,  by  the  last  article,  divisible  by  the  greatest  common  divisor. 

Dividing  the  preceding  divisor  by  the  last  remainder,  a  still 
smaller  remainder  will  be  found,  which  is  divisible  by  the  great 
est  common  divisor;  and  by  continuing  this  process  with  each 
remainder  and  the  preceding  divisor,  quantities  smaller  and 
smaller  are  found,  which  are  all  divisible  by  the  greatest  com 
mon  divisor,  until  at  length  the  greatest  common  divisor  must 
be  obtained.  Hence  we  have  the  following 


o 
RULE. 


Divide  the  greater  quantity  by  the  less,  and  the  preceding  divisor 
by  the  last  remainder,  till  nothing  remains  ;  the  last  divisor  will  be 
the  greatest  common  divisor. 

When  the  remainders  decrease  to  unity,  the  given  quanti 
ties  have  no  common  divisor  greater  than  unity,  and  are  said 
to  be  incommensurable,  or  prime  to  each  other. 

EXAMPLES. 

1.  What  is  the  greatest  common  divisor  of  372  and  246? 


372 
246 

246 

1 

-the  first  remainder. 

126, 

1 

246 
126 
126  1 120,  the  second  remainder. 

120  IT" 

120  |      6,  the  third  remainder. 
120  I  20 
Here  we  have  continued  the  operation  of  division  until  we 


GREATEST  COMMON  DIVISOR.  fc>Q 

obtain  0  for  a  remainder;  the  last  divisor  (6)  is  the  greatest 
common  divisor.  Thus,  246  and  372,  being  each  divided  by  6, 
give  the  quotients  41  and  62,  and  these  numbers  are  prime 
with  respect  to  each  other;  that  is,  have  no  common  divisor 
greater  than  unity. 

2.  What  is  the  greatest  common  divisor  of  336  and  720  ? 

Ans.  48. 

3.  What  is  the  greatest  common  divisor  of  918  and  522  ? 

Ans.  18. 

96.  In  applying  this  rule  to  polynomials  some  modification 
may  become  necessary.     It  may  happen  that  the  first  term  of 
the  arranged  dividend  is  not  divisible  by  the  first  term  of  the 
divisor.     Tfiis  may  arise  from  the  presence  of  a  factor  in  the 
divisor  which  is  not  found  in  the  dividend,  and  may  therefore 
be  suppressed.     For,  since  the  greatest  common  divisor  of  two 
quantities  is  only  the  product  of  their  common  factors,  it  can 
not  be  affected  by  a  factor  of  the  one  quantity  which  is  not 
found  in  the  other. 

We  may  therefore  suppress  in  the  first  polynomial  all  the 
factors  common  to  each  of  its  terms.  We  do  the  same  with 
the  second  polynomial ;  and  if  any  factor  suppressed  is  com 
mon  to  the  two  polynomials,  we  reserve  it  as  one  factor  of  the 
common  divisor  sought. 

But  if,  after  this  reduction,  the  first  term  of  the  dividend, 
when  arranged  according  to  the  powers  of  some  letter,  is  not 
divisible  by  the  first  term  of  the  arranged  divisor,  we  may  mul 
tiply  the  dividend  by  any  monomial  factor  which  will  render  its 
first  term  divisible  by  the  first  term  of  the  divisor. 

This  multiplication  will  not  affect  the  greatest  common  di 
visor,  because  we  introduce  into  the  dividend  a  factor  which 
belongs  only  to  a  part  of  the  terms  of  the  divisor ;  for,  by  sup 
position,  every  factor  common  to  all  the  terms  has  been  sup 
pressed. 

97.  The  preceding  principles  are  embodied  in  the  following 
general 


56  ALGEBRA, 

RULE. 

1.  Arrange  the  two  polynomials  according  to  the  powers  of  some 
letter  ;  suppress  all  the  monomial  factors  of  each ;  and  if  any  fac 
tor  suppressed  is  common  to  the  two  polynomials,  reserve  it  as  one 
factor  of  the  common  divisor  sought. 

2.  Multiply  the  first  polynomial  by  such  a  monomial  factor  as 
will  render  its  first  term  divisible  by  the  first  term  of  the  second 
polynomial ;  then  divide  this  result  by  the  second  polynomial,  and 
continue  the  division  till  the  first  term  of  the  remainder  is  of  a 
lower  degree  than  the  first  term  of  the  divisor. 

3.  Take  the  second  polynomial  as  a  dividend,  and  the  final  re 
mainder  in  the  first  operation  as  a  divisor,  and  proceed  as  before, 
and  so  on  till  a  remainder  is  found  that  will  divide  the  preceding 
divisor.    TJiis  remainder,  multiplied  by  the  common  factors,  if  any, 
reserved  at  the  beginning,  will  give  the  greatest  common  divisor. 

EXAMPLES. 

1.  Find  the  greatest  common  divisor  of  x34-4x24-5x-|-2  and 
ic24-5x-f4. 

x3+4x24-5x-f2 
x34-5x2+4x 


x-1 


—  x24-  x4-2 
_  x2— 5x— 4 


6x4-6 

Suppressing  the  factor  6  in  this  remainder,  we  have  x+I 
for  the  next  divisor. 

x2  4-  5x4-4  1x4-1 
x2-j-  x       |x4-4 
4x4-4 
4x4-4 

Here  the  division  is  exact;  hence,  by  the  rule,  x4-l  is  the 
greatest  common  divisor  sought 

2.  Find  the  greatest  common  divisor  of  6x3— Tax2— 20a2x 
and  6x2+2ax-8a2. 

Suppressing  the  factor  2  in  the  second  polynomial,  we  pro 
ceed  thus : 


GREATEST  COMMON  DIVISOR. 


5? 


6x3-7ax2-20a2x 
2-  Sa2x 


-9ax2-12a2x 
—  9acc2—  3a2a;+12a3 
-  9a2x-12a3 

Suppressing  the  factor  —  3a2, 

ax— 4a2 


—  Sax— 4a2 


— a 


Hence  3#+4a  is  the  greatest  common  divisor. 
3.  Find  the  greatest  common  divisor  of  4a3— 2a2— 3a+l 
and  3a2— 2a— 1. 

We  first  multiply  the  greater  polynomial  by  3,  to  render 
its  first  term  divisible  by  the  first  term  of  the  other  polyno 
mial. 

12a3— 6a2— 9a-f  3  I  3a2— 2a— 1 
12a3— 8a2— 4a        |  4a,  +2 


2a2-5a+3 
6a2-15a+  9 

-lla+11 

Here  we  multiply  the  first  remainder  by  3,  to  render  the 
first  term  divisible  by  the  first  term  of  the  divisor.  As  the 
two  partial  quotients  4a  and  2  have  no  connection,  they  are 
separated  by  a  comma. 

Kejecting  the  factor  —11  from  the  second  remainder,  we 
proceed  as  follows : 

3a2— 3a 


3a+l 


a-l 
a-l 


Hence  a— 1  is  the  greatest  common  divisor. 

4.  Find  the  greatest  common  divisor  of  a2— 3a6-f2&2  and 

a*_a6— 262.  Ans.  a— 25. 

02 


58  ALGEBRA. 

5.  Find  the  greatest  common  divisor  of  a3  —  azb-\-Bab'2—Bb9 
and  a2—  5a5-f452.  Ans.  a-b. 

6.  Find  the  greatest  common  divisor  of  Sx3  —  13cc2-f  23£—  21 
and  6z3+#2-44x+21.  Ans.  3z-7. 

7.  Find  the  greatest  common  divisor  of 

x4_7o;3+8z2+28z-48  and  a3-8x2+19x-14. 

Ans.  x—2. 

98.  To  find  the  greatest  common  divisor  of  three  quantities.  — 
Find  the  greatest  common  divisor  of  the  first  and  second,  and 
then  the  greatest  common  divisor  of  this  result  and  the  third 
quantity.  The  last  will  be  the  greatest  common  divisor  re 
quired. 

EXAMPLES. 

1.  Find  the  greatest  common  divisor  of  3a2m2,  6&2m2,  and 
12ra3ce.  Ans.  3m2. 

2.  Find  the  greatest  common  divisor  of  4x3—  21x2  +  15x+20, 
x2—  6^+8,  and  xz—  x—  12.  Ans.  x—4. 

3.  Find    the    greatest    common    divisor    of    6x4  -\-  x3  —  x^ 
4z3—  6x2—  4x  +  3,  and  2x3  +  cc2+x—  1.  Ans.  2x—  1. 

4.  Find  the  greatest  common  divisor  of  4x4-f  9x3+2x2—  2x—  4, 

—  x+2,  and  x3  +  ^2—  ^+2.  Ans.  a+2. 


"    LEAST  COMMON  MULTIPLE. 

99.  One  quantity  is  a  multiple  of  another  when  it  can  be  di 
vided  by  it  without  a  remainder.    Thus  5ab  is  a  multiple  of  5, 
also  of  a  and  of  b.     When  one  quantity  is  a  multiple  of  an 
other,  the  former  must  be  equal  to  the  product  of  the  latter  by 
some  entire  factor.     Thus,  if  a  is  a  multiple  of  bt  then  a=mb, 
where  m  is  an  entire  number. 

100.  A  common  multiple  of  two  or  more  quantities  is  one 
which  can  be  divided  by  each  separately  without  a  remainder. 
Thus  20a3Z>2  is  a  common  multiple  of  4ab  and  5«2&2. 

101.  The  least  common  multiple  of  two  or  more  quantities  is 
the  least  quantity  that  can  be  divided  by  each  without  a  re- 


LEAST  COMMON   MULTIPLE.  59 

mainder.     Thus  12a2  is  the  least  common  multiple  of  3a2.and 


102.  It  is  obvious  that  the  least  common  multiple  of  two  or 
more  quantities  must  contain  all  the  factors  of  each  of  the  quan 
tities,  and  no  other  factors.  Hence,  when  the  given  quantities 
can  be  resolved  into  prime  factors,  the  least  common  multiple 
may  be  found  by  the  following 

EULE. 

Resolve  each  of  the  quantities  into  its  prime  factors  ;  take  each 
factor  the  greatest  number  of  times  it  enters  any  of  the  quantities; 
multiply  together  the  factors  thus  obtained,  and  the  product  will  be 
the  least  common  multiple  required. 

EXAMPLES. 

1.  Find  the  least  common  multiple  of  9x2y  and  ~L2xyz. 
Eesolving  into  factors,  we  have 

9x2?/  =  3  x  Sxxy,  and  I2xy2  —  3  X  2  x  2xyy. 
The  factor  3  enters  twice  in  the  first  quantity,  also  the  fac 
tor  2  enters  twice  in  the  second;  x  twice  in  the  first,  and  y 
twice  in  the  second.     Hence  the  least  common  multiple  is 
2  x  2  x  3  x  Sxxyy,  or  36x2?/2. 

2.  Find  the  least  common  multiple  of  4a262,  6a26,  and 
We  have  4a262  =  2  x  2aa56, 


=2  x  5aaaxx. 
Hence  the  least  common  multiple  is 

2  X  2  x  3  x  5aaabbxx,  or  60a?bW. 

3.  Find  the  least  common  multiple  of  a2x—  2abx+b2x  and 
y—tyy. 
Here  we  h  ave  a?x  —  2  abx  -f  b2x  =  (a  —b)(a—  b)  x, 


Hence  the  least  common  multiple  is 

(a  —  b)  (a  —  b)  (a  +  b)  xy,  or  a3xy  —  ab2xy  —  a2bxy  +  b3xy. 


60  ALGEBRA. 

4  Find  the  least  common  multiple  of  5a262,  10a&8,  and  2dbx, 

Ans.  lOaWx. 

5.  Find  the  least  common  multiple  of  3a&2,  4ax2,  552x,  and 
6a2x2.  Ans.  60a2b2x2. 

6.  Find  the  least  common  multiple  of  x*—  3^+2  and  x2—  1. 

Ans.  (x  +  I)(x—  l)(x—  2),  or  x3—  2x2—  cc  +  2. 

7.  Find  the  least  common  multiple  of  a3x-|-63a?  and  5a2—  562. 


or      5a4x—  5a3bx-{-5ab3x  —  564ic. 

103.  "When  the  quantities  can  not  be  resolved  into  factors 
by  any  of  the  preceding  methods,  the  least  common  multiple 
may  be  found  by  applying  the  following  principles  : 

If  two  polynomials  have  no  common  divisor,  their  product 
must  be  their  least  common  multiple  ;  but  if  they  have  a  com 
mon  divisor,  their  product  must  contain  the  second  power  of  this 
common  divisor.  Their  least  common  multiple  will  therefore 
be  obtained  by  dividing  their  product  by  their  greatest  com 
mon  divisor.  Hence,  to  find  the  least  common  multiple  of  two 
quantities,  we  have  the  following 

RULE. 

Divide  the  product  of  the  two  polynomials  by  their  greatest  com 
mon  divisor  ;  or  divide  one  of  the  polynomials  by  the  greatest  com 
mon  divisor,  and  multiply  the  other  by  the  quotient. 

EXAMPLES. 

1.  Find  the  least  common   multiple   of  6x2  —  x  —  1   and 


The  greatest  common  divisor  of  the  given  quantities  is  2x— 
1.     Hence  the  least  common  multiple  is 

2)  or 


2.  Find  the  least  common  multiple  of  x3—  1  and  x2-{-x—2. 

Ans.  (x3-I)(x  +  2). 

3.  Find  the  least   common  multiple  of  x3—  9x2+23x—  15 
and  x2—  Sx+T.  Ans.  (x3  —  9o;2+23x-15)(^—  7). 


LEAST  COMMON   MULTIPLE.  61 

104.  When  there  are  more  than  tivo  polynomials,  find  the 
least  common  multiple  of  any  two  of  them ;  then  find  the 
least  common  multiple  of  this  result,  and  a  third  polynomial ; 
and  so  on  to  the  last. 

4.  Find  the  least  common  multiple  of  a2+2a  — 3,  a2—!,  and 
a-1.  Ans.  (a2-!) (a +3). 

5.  Find  the  least  common  multiple  of  4a2-f-l,  4<x2— 1,  and 
2a-l.  Ans.  16a4-l. 

6.  Find  the  least  common   multiple   of  a3— a,  a3+l,  and 
a3-!.  Ans.  a(a6-l). 

7.  Find  the  least  common  multiple  of  (x-\-2a}3,  (x— 2a)3, 
and  sc2-4a2.  Ans.  (#2-4a2)3. 


62  ALGEBRA. 


CHAPTER  VII. 

FRACTIONS. 

105.  A.  fraction  is  a  quotient  expressed  as  described  in  Art, 
71,  by  writing  the  divisor  under  the  dividend  with  a  line  be 
tween  them.    Thus  r  is  a  fraction,  and  is  read  a  divided  by  b. 

106.  Every  fraction  is  composed  of  two  parts:  the  divisor, 
which  is  called  the  denominator,  and  the  dividend,  which  is 
called  the  numerator. 

107.  An  entire  quantity  is  an  algebraic  expression  which  has 
no  fractional  part,  as  a2  —  2a&. 

An  entire  quantity  may  be  regarded  as  a  fraction  whose  de 
nominator  is  unity.  Thus,  a2  =  y. 

108.  A  mixed  quantity  is  an  expression  which  has  both  en 
tire  and  fractional  parts.     Thus  «2  +  -  is  a  mixed  quantity. 

109.  General  Principles  of  Fractions. —  The  following  princi 
ples  form  the  basis  of  most  of  the  operations  upon  fractions : 

1st.  In  order  to  multiply  a  fraction  by  any  number,  we  must 
multiply  its  numerator  or  divide  its  denominator  by  that  number. 

Thus  the  value  of  the  fraction  —^  is  b.  If  we  multiply  the 
numerator  by  a,  we  obtain  — ,  or  ab  •;  and  if  we  divide  the  de 
nominator  of  the  same  fraction  by  a,  we  obtain  also  ab;  that  is, 
the  original  value  of  the  fraction,  b,  has  been  multiplied  by  a. 

Zd.  In  order  to  divide  a  fraction  by  any  number,  we  must  divide 
its  numerator  or  multiply  its  denominator  by  that  number. 

Thus  the  value  of  the  fraction  ~  is  ab.    If  we  divide  the 


FRACTIONS.  63 

numerator  by  a,  we  obtain  ^,  or  b  ;  and  if  we  multiply  the  de 
nominator  of  the  same  fraction  by  a,  we  obtain  ^  ,  or  b  ;  that 
is,  the  original  value  of  the  fraction  ab  has  been  divided  by  a. 

3d  The  value  of  a  fraction  is  not  changed  if  we  multiply  or  di 
vide  both  numerator  and  denominator  by  the  same  number. 

ab     abm     abmx     T 

Thus.  —  =  -  =  --  =  b. 

a      am      amx 

110.  The  proper  Sign  of  a  Fraction.  —  Each  term  in  the  numer 
ator  and  denominator  of  a  fraction  has  its  own  particular  sign, 
and  a  sign  is  also  written  before  the  dividing  line  of  a  fraction. 
The  relation  of  these  signs  to  each  other  is  determined  by  the 
principles  already  established  for  division.  The  sign  prefixed 
to  the  numerator  of  a  fraction  affects  merely  the  dividend  ;  the 
sign  prefixed  to  the  denominator  affects  merely  the  divisor; 
but  the  sign  prefixed  to  the  dividing  line  of  a  fraction  affects 
the  quotient.  The  latter  sign  may  be  called  the  apparent  sign 
of  the  fraction,  while  the  real  sign  of  the  fraction  is  the  sign 
of  its  numerical  value  when  reduced. 

The  real  sign  of  a  fraction  depends  not  merely  upon  its  <7/> 
parent  sign,  but  also  upon  the  signs  of  the  numerator  and  de 
nominator.  From  Art.  73,  it  follows  that 

ab     —  ab 

-ap  -  =  +&, 

a       —a 

,  —  ab      ab 

and  •  -  =  —  =—b. 

a        -^-a 

Also,  since  a  minus  sign  before  the  dividing  line  of  a  frac 
tion  shows  that  the  quotient  is  to  be  subtracted,  which  is  done 
by  changing  its  sign,  it  follows  that 


ab  _      —ab  _ 
a~       —  a~ 


and 


a  —  a 

Hence  we  see  that  of  the  three  signs  belonging  to  the  numer- 


64  ALGEBRA. 

ator,  denominator,  and  dividing  line  of  a  fraction,  any  two  may 
be  changed  from  -+-  to  — ,  or  from  —  to  -f->  without  affecting  the 
real  sign  of  the  fraction. 

111.  When  the  numerator  or  denominator  of  a  fraction  is  a 
polynomial,  it  must  be  observed  that  by  the  sign  of  the  numer* 
ator  is  to  be  understood  the  sign  of  the  entire  numerator,  as  dis 
tinguished  from  the  sign  of  any  one  of  its  terms  taken  singly. 

m,                 a  +  b  +  c             .     ,                  —a—b—c 
Thus,      — is  equivalent  to  -\ — . 

00  00 

When  no  sign  is  prefixed  either  to  the  terms  of  a  fraction  or 
to  its  dividing  line,  plus  is  always  to  be  understood. 

Reduction  of  Fractions. 

112.  To  reduce  a  Fraction  to  its  Lowest  Terms. — A  fraction  Is 
in  its  lowest  terms  when  the  numerator  and  denominator  con 
tain  no  common  factor ;  and  since  the  value  of  a  fraction  ia 
not  changed  if  we  divide  both  numerator  and  denominator 
by  the  same  number  (Art.  109),  we  have  the  following 

RULE. 

Divide  both  numerator  and  denominator  by  their  greatest  com 
mon  divisor. 

Or,  Cancel  all  those  factors  which  are  common  to  both  numerator 
and  denominator. 

EXAMPLES. 

azbc 
1.  Eeduce  -=-      to  its  lowest  terms. 


a2bc       c  x  a2b 
We  have  ~  070  = 


x  azb' 

Canceling  the  common  factors  a25,  we  have 

c 


CX  \  X2 

2.  Reduce    „        9     to  its  lowest  terms. 


have 


x(c-\-x)      x 

= 


FRACTIONS.  65 

~7a&  7a 

3.  Eeduce  ?   -  ^r-  to  lts  lowest  terms.  Arcs.  ^-. 

lOac  —  ooc  oc 

£.2  _  tt2  i 

4.  Eeduce  —.  -  ,  to  its  lowest  terms.  Ans.  —  -  «. 

a4—  a4  85*+  a* 

2x2—16x  _  6 

5.  Kedace      .2_94a;—  9  tO  itS  lowest  terms- 


6.  Eeduce  0  ?   ^  --  7^3  to  lts  lowest  terms. 


— 

7.  Eeduce  -5  —  ^  ,     ,2  to  its  lowest  terms. 

8.  Eeduce  -^  —  ^  -  5  to  its  lowest  terms. 

a2— 


x2  —  16 
9.  Eeduce  -5—     —  ^  to  its  lowest  terms. 

x2—  cc—  20 

3cc3-16x2-f-23x-6  .     .     , 

10.  Eeduce  ^-^  —  ^   ,  .  1f7  -  5  to  its  lowest  terms. 

—  6 


_  __    . 

11.  Eeduce  ~-^   —  ^—    —5  to  its  lowest  terms. 

JjOC   —  3C    —  X-f-  2i 


—  3 

12.  Eeduce  75-=  —  g  .   '    c—  —  r  to  its  lowest  terms. 
3x3  +  5x2  — 


113.  To  reduce  a  Fraction  to  an  Entire  or  Mixed  Quantity.  —  • 
When  any  term  of  the  numerator  is  divisible  by  some  term  in 
the  denominator,  the  division  indicated  by  a  fraction  may  be 
at  least  partially  performed.  Hence  we  have  the  following 

RULE. 

Divide  the  numerator  ly  the  denominator,  continuing  the  opera 
tion  as  far  as  possible  ;  then  write  the  remainder,  if  any,  over  ihi 
denominator,  and  annex  the  fraction  thus  formed  to  the  entire  part* 

EXAMPLES. 

ax     x^ 

1.  Eeduce  --  to  an  entire  quantity. 

7         O2 

2.  Eeduce  -  =  -  to  a  mixed  quantity. 


66  ALGEBRA. 

3.  Reduce  to  a  mixed  quantity.    Ans.  a-f  x-\ . 

a  — x  a—x 

4.  Reduce  -  —2.  to  an  entire  quantity. 

x—y 

5.  Reduce  —  —  to  a  mixed  quantity. 

ox 

6.  Reduce  -       — ^r—     —  to  a  mixed  quantity. 

7.  Reduce = — ~ o to  an  entire  quantity. 

x2  —  3x+2  J 

8.  Reduce  -    — 2     9  ,     ,2 — '  -  to  an  entire  quantity. 


114.  To  reduce  a  Mixed  Quantity  to  the  Form  of  a  Fraction.  — 
This  problem  is  the  converse  of  the  last,  and  we  may  proceed 
by  the  following 

RULE. 

Multiply  the  entire  part  by  the  denominator  of  the  fraction  ;  to 
the  product  add  the  numerator  with  its  proper  sign,  and  write  the 
result  over  the  denominator. 

EXAMPLES. 

y-y2  _  ^  rrQ  yy* 

1.  Reduce  x-\  --  to  the  form  of  a  fraction.     Ans.  —  . 

x  x 

rtnf  _L  nr*& 

2.  Reduce  x+-     —  to  the  form  of  a  fraction. 

2a  Sax+x* 

Ans.  —  n  -  . 

2x-7  2a 

3.  Reduce  5H  —  5  —  to  the  form  of  a  fraction. 

ox 

/v»  _        /y       _.  "1 

4.  Reduce  lH  —          -  to  the  form  of  a  fraction. 


x—3 
--r 
ox 

_  gc2 
6.  Reduce  7-\  —  5  —  jj-  to  the  form  of  a  fraction. 


x—3 

5.  Reduce  1-f  2#-f--r  —  to  the  form  of  a  fraction 
ox 


FRACTIONS.  67 

7.  Reduce  2a—  7  —    ~  „    to  the  form  of  a  fraction. 


8   Eeduce  (a—  I)2—  -  -  '-  to  tlie  form  of  a  fraction. 

A       (a- 
Ans.  - 


115.  To  reduce  Fractions  having  Different  Denominators  fo 
Equivalent  Fractions  having  a  Common  Denominator  : 

*          ct    (*  rrn 

Suppose  it  is  required  to  reduce  the  fractions  7,  -3,  and  —  to 

a  common  denominator.  Since,  by  Art.  109,  both  terms  of  a 
fraction  may  be  multiplied  by  the  same  quantity  without 
changing  its  value,  we  may  multiply  both  terms  of  each  frac 
tion  by  the  product  of  the  denominators  of  the  other  fractions, 
and  we  shall  have 

a_adn    c  _bcn         ,   m__bdm 

b     bdn   d     bdn  n      bdn' 

The  resulting  fractions  have  the  same  value  as  the  proposed 
fractions,  and  they  have  the  common  denominator  bdn.  Hence 
we  have  the  following 

BULE. 

Multiply  each  numerator  into  all  the  denominators,  except  its 
own,  for  a  new  numerator,  and  all  the  denominators  together  for 
the  common  denominator. 

EXAMPLES. 

1.  Beduce   T   and   -      -   to  equivalent  fractions  having  a 

0  C 

,  A       ac    a 

Common  denominator.  Ans.  y-, 


,  . 

be        be 

2.  Eeduce  ~-,  ~-,  and  -  to  equivalent  fractions  having  a 
common  denominator. 

o      0,-v,  A™. 

3.  Reduce  p  -jr-j  and  a+-^-  to  equivalent  fractions  having 
a  common  denominator. 


68  ALGEBRA. 

4.  Keduce  ->  -y»  and  -     -  to  equivalent  fractions  having 

a  common  denominator. 

x  x-\-~L        ,  1  —  x 

5.  Keduce  ->  —  ^—  ,  and  —  —  to  equivalent  fractions  hav« 

o         O  1  -\-X 

ing  a  common  denominator. 

116.  Fractions  may  always  be  reduced  to  a  common  denom 
inator  by  the  preceding  rule  ;  but  if  the  denominators  have  any 
common  factors,  it  will  not  be  the  least  common  denominator. 
The  least  common  denominator  of  two  or  more  fractions  must 
be  the  least  common  multiple  of  their  denominators. 

Suppose  it  is  required  to  reduce  the  fractions  ^  and  -:-  to 

ox  *±x 

equivalent  fractions  having  the  least  common  denominator. 
The  least  common  multiple  of  the  denominators  is  12#2.    Mul- 


tiply  both  terms  of  the  first  fraction  by  -0-5-,  or  4,  and  both 

12x2 
terms  of  the  second  fraction  by  --  ,  or  3x,  and  we  shall  have 

4:X 

Sa          ,    I5bx 
{     l 


which  are  equivalent  to  the  given  fractions,  and  have  the  least 
common  denominator.     Hence  we  deduce  the  following 

RULE. 

Find  the  least  common  multiple  of  all  the  denominators,  and  use 
this  as  the  common  denominator. 

Divide  this  common  denominator  by  each  of  the  given  denomina 
tors  separately,  and  multiply  each  numerator  by  the  corresponding 
quotient.  The  products  will  be  the  new  numerators. 

6.  Keduce  ^-  and  -^^  to  equivalent  fractions  having  the 

4ac  x 

least  common  denominator.  Ans.  --    and 


_  j 

7.  Reduce  -  —  7  and    °     19  to  equivalent  fractions  having 

a  —  b  az  —  b2  ,       7N2  , 

A        (a  +  b)2        ,    c+d 

the  least  common  denominator.         Ans.  ~  —  =£•  and  -5  —  ^. 

a2  —  62  az—b2 


FRACTIONS.  69 

8.  Eeduce  4»  —  »  and  -  to  equivalent  fractions  having  the 

x3  x2  x 

least  common  denominator. 

9a      7b      lla         ,  7(a+5)  .  .     ,     .  f 

9.  Eeduce  ?—  ,  ,7^-,  OQ—  »  and  -—-1  to  equivalent  frac- 

$7n   86m    28m  4m 

tions  having  the  least  common  denominator. 
o        g  2x  _  3 

10.  Reduce  -,  ^  -  7,  and  -r-z  —  T  to  equivalent  fraction^ 

cc    2cc—  1  4cc2—  1 

having  the  least  common  denominator. 

r^ 

Addition  of  Fractions. 

117.  The  denominator  of  a  fraction  shows  into  how  many 
parts  a  unit  is  to  be  divided,  and  the  numerator  shows  how 
many  of  those  parts  are  to  be  taken.  Fractions  can  only  be 
added  when  they  are  like  parts  of  unity  ;  that  is,  when  they 
have  a  common  denominator.  In  that  case,  the  numerator  of 
each  fraction  will  indicate  how  many  times  the  common  frac 
tional  unit  is  repeated  in  that  fraction,  and  the  sum  of  the  nu 
merators  will  indicate  how  many  times  this  result  is  repeated 
in  the  sum  of  the  fractions.  Hence  we  have  the  following 

RULE. 

Reduce  the  fractions  to  a  common  denominator  /  then  add  the 
numerators  together,  and  write  their  sum  over  the  common  denomi 
nator. 

If  there  are  mixed  quantities,  we  may  add  the  entire  and 
fractional  parts  separately. 

EXAMPLES. 

/JT*  IT  o 

1.  What  is  the  sum  of  o  and  gf 

Reducing  to  a  common  denominator,  the  fractions  become 


jX 

Adding  the  numerators,  we  obtain  —  . 

It  is  plain  that  three  sixths  of  x  and  two  sixths  of  x  mako 
five  sixths  of  x. 


70  ALGEBRA. 


2.  What  is  the  sum  of  T,  -,  and  —  ? 

o   a  n  7       ,         ,  7 

.       aan  +  ben  4-  ft#m 

J.715. 

, 

3.  What  is  the  sum  of  -  7  and 


a-j-ft  a—b 

t       TTTT-     A    '       ^  £     K  %a  J     a  0 

4.  What  is  the  sum  of  5x.  -^—^j  and  —  -.  -  ? 

'  Sxz  4x 

5.  What  is  the  sum  of  2a,  3a+->  and  &+•• 


.   6a-|--r=-. 

45 

6.  What  is  the  sum  of  a+x,  -  ,  and  —  ^-  ? 

a-x  a  2 

Ans. 


a  —  b  a—  ax 

7.  What  is  the  sum  of       —  and  ^~-  ?  Ans.  a. 

2i  A 

o   -rrn    ^  •    xi  r  a    «—  2m         ,   a+2m0 

8.  What  is  the  sum  of  ^,  —  j  —  ,  and  —  I—  —  ? 

n   TTTI,  4.  •    4.1  r  ma—b        •>  na-\-b~ 

9.  What  is  the  sum  of  -  and  --  —  ? 

m-\-n  m-\-n 

•in   TTTI,  4  -    xi.  c     x—n          y—z  .     2z+n   „ 

10.  What  is  the  sum  of  —       —  ,  —  —    —  ,  and  —  —  —  ? 

x+y+z    x+y  +  z  x+y+z 

11.  What  is  the  sum  of         =     and 


--  -  ,          - 

12.  What  is  the  sum  of  -=-.  -  ^-,  —  -=-,  -  j-,  and  —-=-7  -  -^ 
b(a—b)         5(a—  ft)  5(a—  ft) 

.  9. 


r     +  aj      —  a;         —  c 

13.  What  is  the  sum  of  -  --  ,  ——  ,  --  1  ,         ,  --  T—--I 
1—  a;  1  +  x          1+x2  1—  a;3 

and  —  1?  -r 

Subtraction  of  Fractions. 

118.  Fractions  can  only  be  subtracted  when  they  are  like 
parts  of  unity  ;  that  is,  when  they  have  a  common  denomina 
tor.  In  that  case,  the  difference  of  the  numerators  will  indi 
cate  how  many  times  the  common  fractional  unit  is  repented 
in  the  difference  of  the  fractions.  Hence  we  have  the  following 


FRACTIONS.  71 

BULK 

Reduce  the  fractions  to  a  common  denominator  ;  then  subtract 
Hie  numerator  of  the  subtrahend  from  the  numerator  of  the  minu 
end,  and  write  the  result  over  the  common  denominator. 

EXAMPLES. 

2x      .        ,  Sx 
1.  From  -£-  subtract  —  . 
o  o 

Beducing  to  a  common  denominator,  the  fractions  become 


. 
15-  and  IS" 

IQx    9x     x 
Hence  we  nave  -jg  —  15  "15"' 

and  it  is  plain  that  ten  fifteenths  of  cc,  diminished  by  nine  fif 
teenths  of  x,  equals  one  fifteenth  of  x. 

Sx 


2.  From  -=-  subtract 

/  o 

9a—  4x      ,  5a—  Bx 

3.  From  —  =  —  subtract  --  5  —  . 

/  o 

It  must  be  remembered  that  a  minus  sign  before  the  divid 
ing  line  of  a  fraction  affects  the  quotient  (Art.  Ill);  and  since 
a  quantity  is  subtracted  by  changing  its  sign,  the  result  of  the 
subtraction  in  this  case  is 

9a—  4x     5a  —  3x 
~T~       ~3~; 

which  fractions  may  be  reduced  to  a  common  denominator, 
ai>d  the  like  terms  united  as  in  addition. 

ax        ,  ax  2acx 

4.  From  7  —  subtract    —  -.  Ans. 

b—  c 


,    ,-,         0    ,  2  +  7x  5x-6  355.T-6 

5.  From  2x-\  --  5  —  subtract  x  --  ^  —  .       Ans. 


21  168 


/v»  />*  ___  ,  / 

6.  From  8^+^:7  subtract  cc 


Zo  G 

a-\-b  a—o 

7.  From  -~-  subtract  -—  . 


72  ALGEBKA. 


,  -  .        t       - 

8.  From  -  r  —  subtract  —  -  —  .  Ans. 


6o-76      ,  .      .  5a  64a*-15a2-63i2 

-  From  3^=25  SubtraCt  95' 


10.  From      ,+  /    subtract  unity. 

4:dO 

11.  From  —^-  subtract  =- — -^. 

lly  Ix—by 

12.  From  —?—  subtract  -^—n. 


v*    — *  r| 

Multiplication  of  Fractions. 
119.  Let  it  be  required  to  multiply  ^  by  ~. 
First  let  us  multiply  |  by  c.     According  to  the  first  princi 
ple  of  Art.  109,  the  product  must  be  T\ 

But  the  proposed  multiplier  was  ^ ;  that  is,  we  have  used  a 

d 

multiplier  d  times  too  great.  We  must  therefore  divide  the 
result  by  d;  and,  according  to  the  second  principle  of  Art. 
109,  we  obtain 

CtO  '       .  i       ,     •        Uj  .       L        CiC 

r-;,"  that  is,  rx~7— r~7« 
Jc?  &     d    bd 

Hence  we  have  the  following 

RULE. 

Multiply  the  numerators  together  for  a  new  numerator,  and  the 
denominators  for  a  new  denominator. 

Entire  and  mixed  quantities  should  first  be  reduced  to  frac« 
tional  forms.  Also,  if  there  are  any  factors  common  to  the  nu 
merator  and  denominator  of  the  product,  they  should  be  can. 
celed. 

EXAMPLES. 

_..      .        x  ,      2cc  x2 

1.  Multiply  g  by  --.  Ans.      . 

x  , 

2.  Multiply  -  by 


FRACTIONS.  73 


bx  ,       a 
3.  Multiply  6  +  —  by  -. 


_ 

4.  Multiply  --  by 


a  a_ 

5.  Multiply  ^-^  by  —  ^. 

of    4x 

6.  Multiply  together        --,  and 


7.  Multiply  together  — ,  — ,  and 

cc-f  1 

8.  Multiply  together  x,  ~    — ,  and 

ct 

v  /j/y^Y* 

9.  Multiply  s  by 


10.  Multiply  463c6Ua763cM  ^  a2&2c*- 

11.  Multiply  together  ^^,  ^^,  -^^  and  — . 

Ans.  — XT. 


a- 


-  , 

12.  Multiply  together        --  ,  =    —     '  and 
7(m  —  ?ij    39  (a—  b) 


3dn     3bm     5mn  ,        \\abc 
13.  Multiply  ---  1  —  =--—  TH—  by 
r  J    4ac        7         66c 


a2—  cc2    bc  +  bx         ,    c— 

14.  Multiply  together  —  :—,  —  -  -,  —  -  ,  ana 
r  J  22      2 


c—x     a2^-ax  a  —  x 


_^  _y 

15.  Multiply  together  -     —  ,  -  -~-~,  and 
1-f?/    x  +  x2 


A 
Ans.  —. 

4y 


ic    nr  i.-  i 
16.  Multiply 
rJ 


17  .  - 

17.  Multiply  —  —  ^-j  —  r.  by  -r-  —  7.  -dn*. 

1  J       —  2     J  a2  +  ab  a 

T) 


74  ALGEBRA. 

18.  Multiply  xz-x+ 1  by  ^4-jfl         Ans. 

-in  TVT  u-  i     a  +  ^     a~ b       462  a  4-6 

19.  Multiply  —----          by  --.  An,.  2. 


20.  Multiply  V--+1  by  -9-f--  +  l. 

a2     a  ;   a2     a  a*  '  a*  ' 

120.  Multiplication  of  Quantities  affected  with' Negative  Expo 
nents. — Suppose  it  is  required  to  multiply  -^  by  — . 

According  to  the  preceding  article,  the  result  must  be  — . 

1  1  a 

But,  according  to  Art.  76,  -3  may  be  written  a~3 ;   —  may  be 
-»  a  a 

written  a~2;   and  -g  may  be  written  a~5. 

Hence  we  see  that     a~3xa~2=ia-s; 

that  is,  the  rule  of  Art.  58  is  general,  and  applies  to  negative  as 
well  as  positive  exponents. 

EXAMPLES. 

1.  Multiply  —x~2  by  x~\  Ans.  — x~5,  or  —  — 

2.  Multiply  a-2  by  -a3. 

3.  Multiply  a~3  by  a3. 

4.  Multiply  a~m  by  an. 

5.  Multiply  «x~m  by  a~n. 

6.  Multiply  (a— 6)5  by  (a— b)~3. 

Division  of  Fractions. 

121.  If  the  two  fractions  have  the  same  denominator,  then 
the  quotient  of  the  fractions  will  be  the  same  as  the  quotient 
of  their  numerators.     Thus  it  is  plain  that  f  is  contained  in  -J 
as  often  as  3  is  contained  in  9.     If  the  two  fractions  have  not 
the  same  denominator,  we  may  perform  the  division  after  hav 
ing  first  reduced  them  to  a  common  denominator.     Let  it  be 

required  to  divide  j  by  -, 


FRACTIONS.  75 

Eeducing  to  a  common  denominator,  we  have  ^  to  be  di 
vided  by  j-|.  It  is  now  plain  that  the  quotient  must  be  repre- 
sented  by  the  division  of  ad  by  be,  which  gives  j-  ; 

a  result  which  might  have  been  obtained  by  inverting  the 
terms  of  the  divisor  and  multiplying  by  the  resulting  frac 
tion  ;  that  is.  ,       , 
a^€__a_    d_ad 

b  '  d~ b     c~bc 
Hence  we  have  the  following 

RULE. 

Invert  the  terms  of  the  divisor,  and  multiply  the  dividend  by  the 
resulting  fraction. 

Entire  and  mixed  quantities  should  first  be  reduced  to  frac 
tional  forms. 

EXAMPLES. 
X  Zx 

1.  Divide  ^  by  -^-.  Ans.  1J. 

2.  Divide  -=-  by  -7. 

b      J    d 

8'     Divide    7^3    bJ 


,    -p..   .,     x+1         Zx 
L  Divide  — ^—  by  -^-. 

D  O 

•    TV   -j     x—b  .      3cx 

).  Divide  -^-7-  by  -rr.  Ans. 


6.  Divide  ~T  '7    by 


C  —  X 

a 


7-  Divide  ^b  +  ^Tb  ^  l^b-aTV  Ans'  Unit^ 


8.  Divide  7a2-3cc+-  by  &2--. 
n     J          3 


—  an 


76  ALGEBRA, 

,.    T>..   . ,  3a6  ..  a — b 

9.  Divide  15x2—  -=—  by  x .          Ans.  = = =7. 

5c      J  c  5cx— 5a  +  56 

ce3  a6  ax2 — < 

10.  Divide  x2-{ 7  by  7  —  ^.          Ans. 

a—b     J   a—b 

27  (a- 6) 


11.  Divide  ^         — -/  by     OQ    .     -^r. 

32(m+6)     J    128w(m4-6) 

12.  Divide  ^T+-  by  -9 h-.  

y*      x     >   y2      y      x  y 

13.  Divide  -+-  by .  Ans.  Unity. 


1  1 

14  Divide  x2-f-+2  by  a-f— 

X2  CC 


15.  Divide 

'    a+b- 

Ans.  a2— 62+c2  — 2ac. 


Divide  a2-62-c2-26c  by      ^ 
J   a+6— c 


122.  Division  of  Quantities  affected  with  Negative  Exponents. — 

Suppose  it  is  required  to  divide  —s  by  -3.     According  to  the 

preceding  article,  we  have 

1      a3      a3      1 


But,  according  to  Art.  76,  —5  may  be  written  a~5  ;    -3  may  be 

written  a~3  :  and  —  may  be  written  a~2.    Hence  we  see  that 

a2       J 


that  is,  the  rule  of  Art.  72  is  general,  and  applies  to  negative  as 
well  as  positive  exponents. 

EXAMPLES. 

1.  Divide  or5  by  —  a~2.  Ans.  —  or3,  or  —  ^ 

2.  Divide  —a2  by  a~l. 

3.  Divide  1  by  a~4. 

4.  Divide  6«n  by  —  2a~3. 


FRACTIONS.  77 

5.  Divide  bm~n  by  bm. 

6.  Divide  12x-2y~4  by  —  4xy2. 

7.  Divide  (x—  y)~4  by  (x—  ?/)-6. 

123.  The  Reciprocal  of  a  Fraction.  —  According  to  the  defini 
tion  in  Art.  34,  the  reciprocal  of  a  quantity  is  the  quotient  aris 
ing  from  dividing  a  unit  by  that  quantity.  Hence  the  recipro 

cal  of  \  is  .  a  b_bf 

D  JL  -7-  7  —  IX  —  —  —  / 

b  a     a' 

that  is,  the  reciprocal  of  a  fraction  is  the  fraction  inverted. 

a  r)  1  o^ 

Thus  the  reciprocal  of  ^  -  is  -  ;  and  the  reciprocal  of 
_!  b+x          a 

is  b+°: 


It  is  obvious  that  to  divide  by  any  quantity  is  the  same  as  to 
multiply  by  its  reciprocal,  and  to  multiply  by  any  quantity  is  the 
same  as  to  divide  by  its  reciprocal. 

124.  How  to  simplify  Fractional  Expressions.  —  The  numerator 
or  denominator  of  a  fraction  may  be  itself  a  fraction  or  a  mixed 

quantity,  as  -j-.     In  such  cases  we  may  regard  the  quantity 

above  the  line  as  a  dividend,  and  the  quantity  below  it  as  a 
divisor,  and  proceed  according  to  Art.  121. 

Thus,  24-r.£=ix£=Y=3£. 

The  most  complex  fractions  may  be  simplified  by  the  appli 
cation  of  similar  principles. 

EXAMPLES. 

1+f 

1.  Simplify  the  fraction  -  2 

!+! 

b+a     a+b 
This  expression  is  equivalent  to  —  -,  —  •  --  —  » 

or  to  —  r-X—  —  T'  which  is  equal  to  r,  Ans. 

b       a     b  o 


78  ALGEBRA. 

i a— 

2.  Simplify  _£=!2.  Am.  -a-±^. 

^ a_  a—m 

a+m 

24 

3,  Find  the  value  of  the  fraction  H.  Ans.  2. 


4.  Find  the  value  of  the  fraction    ,  f^ 

lla& 

a+b     a—b 

~    ci-       f/.     c-4-d     c — d  ac — bd 

5.  Simplify  — —= r.  Ans. j-,, 

^-h^     a— 6  ac+6^ 

c—dc+d 

a+x    a—x 

„    „.      ,.«     a—x     a+x 

6.  Simplify  -  — .  Ans. 

a+x     a—x 

a—x     a+x 


—  m 


n  mz — n2 


1- 

n     m 


7.  Simplify  — ^ ^ x~3~i — §•  Ans.  m. 


a 

8.  Simplify  -      -7-  adn  +  am 

b  -i Ans. 


bdn+bm+cn 


EQUATIONS  OF  THE   FIRST   DEGREE.  1\) 


CHAPTEK  YIII. 

EQUATIONS    OF    THE    FIRST    DEGREE. 

125.  An  equation  is  an  expression  of  equality  between  two 
algebraic  quantities.     Thus  Sx=2ab  is  an  equation  denoting 
that  three  times  the  quantity  x  is  equal  to  twice  the  product 
of  the  quantities  a  and  b. 

126.  The  first  member  of  the  equation  is  the  quantity  on  the 
left  side  of  the  sign  of  equality,  and  the  second  member  is  the 
quantity  on  *the  right  of  the  sign  of  equality.      Thus,  in  the 
preceding  equation,  Sx  is  the  first  member,  and  2ab  the  second 
member. 

127.  The  two  members  of  an  equation  are  not  only  equal 
numerically,  but  must  have  the  same  essential  sign.     If,  in  the 
preceding  equation,  x  represents  a  negative  quantity,  then  the 
first  member  is  essentially  negative,  and  the  second  member 
must  also  be  negative ;  that  is,  either  a  or  b  must  represent  a 
negative  quantity. 

128.  Equations  are  usually  composed  of  certain  quantities 
which  are  known,  and  others  which  are  unknown.     The  known 
quantities  are  represented  either  by  numbers,  or  by  the  first 
letters  of  the  alphabet;   the  unknown  quantities  are  usually 
represented  by  the  last  letters  of  the  alphabet. 

129.  A  root  of  an  equation  is  the  value  of  the  unknown, 
quantity  in  the  equation ;  or  it  is  any  value  which,  being  sub 
stituted  for  the  unknown  quantity,  will  satisfy  the  equation. 
For  example,  in  the  equation 

See— 4  =  24— a, 

suppose  x  —  7.    Substituting  7  for  cc,  the  first  member  becomes 
3x7—4;   that  is,  21—4,  or  17;   and  the  second  member  be- 


80  ALGEBRA. 

comes  24  —  7;  that  is,  17.  Hence  7  is  a  root  of  the  equation, 
because  when  substituted  for  x  the  two  members  are  found  to 
be  equal. 

130.  A  numerical  equation  is  one  in  which  all  the  known 
Quantities  are  represented  oy  figures;  as,  x3-f  4x2  —3x4-12. 

131.  A  literal  equation  is  one  in  which  the  known  quantities 
are  represented  by  letters,  or  by  letters  and  figures. 

Thus  x3  +  axz  +  bx  =  m.       ) 

and   rf-sox'fc^      are  llteral  e(iuatlons- 


132.  The  degree  of  an  equation  is  denoted  by  the  greatest 
number  of  unknown  factors  occurring  in  any  term. 

If  the  equation  involves  but  one  unknown  quantity,  its  de 
gree  is  denoted  by  the  exponent  of  the  highest  power  of  this 
quantity  in  any  term. 

If  the  equation  involves  more  than  one  unknown  quantity, 
its  degree  is  denoted  by  the  greatest  sum  of  the  exponents  of 
the  unknown  quantities  in  any  term. 

Thus  ax-\-b  =  cx+d  is  an  equation  of  tine  first  degree,  and  is 
sometimes  called  a  simple  equation. 

4:X2—2x=5—x2  and  7xy—  4x-f-?/=40  are  equations  of  the 
second  degree,  and  are  frequently  called  quadratic  equations. 

x3  +  ax2  =  2b  and  x'2-{-%xy'*+y  =  m  are  equations  of  the  third 
degree,  and  are  frequently  called  cubic  equations. 

So  also  we  have  equations  of  the  fourth  degree,  sometimes 
called  bi-quadratic  equations;  equations  of  the  fifth  degree,  etc., 
up  to  the  nth  degree. 

Thus  xn  +  axn~l  —  b  is  an  equation  of  the  nth  degree. 

133.  To  solve  an  equation  is  to  find  the  value  of  the  unknown 
quantity,  or  to  find  a  number  which,  being  substituted  for  the 
unknown  quantity  in  the  equation,  renders  the  first  member 
identical  with  the  second. 

The  difficulty  of  solving  equations  depends  upon  their  de 
gree,  and  the  number  of  unknown  quantities  they  contain. 


EQUATIONS  OF  THE   FIRST  DEGREE.  81 

134.  Axioms.  —  The  various  operations  which  we  perform 
upon  equations,  in  order  to  deduce  the  value  of  the  unknown 
quantities,  are  founded  upon  the  following  principles,  which 
are  regarded  as  self-evident. 

1.  If  to  two  equal  quantities  the  same  quantity  be  added,  the 
sums  will  be  equal. 

2.  If  from  two  equal  quantities  the  same  quantity  be  sub 
tracted,  the  remainders  will  be  equal. 

3.  If  two  equal  quantities  be  multiplied  by  the  same  quanti 
ty,  the  products  will  be  equal. 

4.  If  two  equal  quantities  be  divided  by  the  same  quantity, 
the  quotients  will  be  equal. 

135.  Transposition.  —  Transposition  is  the  process  of  changing 
a  term  from  one  member  of  an  equation  to  the  other  without 
destroying  the  equality  of  the  members. 

Let  it  be  required  to  solve  the  equation 

x-\-a=b. 

If  from  the  two  equal  quantities  x-{-a  and  b  we  subtract  the 
same  quantity  a,  the  remainders  will  be  equal,  according  to  the 
last  article,  and  we  shall  have 

x+a—  a  =  b  —  a, 
or  x=b  —  a. 

Let  it  be  required  to  solve  the  equation 

x—  a  —  b. 

If  to  the  two  equal  quantities  x—a  and  b  the  same  quantity 
a  be  added,  the  sums  will  be  equal,  according  to  the  last  arti 
cle,  and  we  have  x  —  a  +  #—  b-\-a, 

or  x 


136.  Hence  we  perceive  that  we  may  transpose  any  term  of  an 
equation  from  one  member  of  the  equation  to  the  other,  provided  we 
change  its  sign. 

It  is  also  evident  that  we  may  change  the  sign  of  every  term  of 
an  equation  without  destroying  the  equality  ;  for  this  is,  in  fact,  the 
same  thing  as  transposing  every  term  in  each  member  of  the 
equation. 

D2 


82  ALGEBRA. 

EXAMPLES. 

In  the  following  examples,  transpose  the  unknown  terms  to 
the  first  member  and  the  known  terms  to  the  second  member. 

1.  5z+12  =  3x+18.  Am.  6x~  Bx=lS  -12. 

2.  4x—  7  =  21  —  Sx.  Ans.  4x  +  3x=2l+7. 

3.  2x—  15  =  —  7x+30.  Ans.  2z-f  7x=30+15. 

4.  ax+bc=m—  2x.  Ans.  ax+2x=m—bc. 
6.  4ax—  b+2c=3x—  2ab—  Bmx. 

Ans.  4ax—  3x  -\-3mx  =b—2c—2ab. 

6,  4ab—  ax—  2c=bx—  3m.       Ans.  ax+bx=4:ab—2c+3m. 

7.  ab—cx  —  2mx=3ax  —  4b. 

Ans. 


137.  To  clear  an  Equation  of  Fractions.  —  Let  the  equation  be 
?=&  If  we  multiply  each  of  the  equal  quantities  -  and  b  by 

the  same  quantity  a,  the  products  will  be  equal  by  Art.  134, 
and  we  shall  have  x~ab. 

,    x    x 
Suppose  the  equation  is  —  \-j-=m. 

If  we  multiply  each  of  the  members  of  the  equation  by  a,  we 

shall  have  ax 

x-\  —  T-=am. 
o 

If  we  multiply  each  of  the  members  of  this  equation  by  6, 
We  shall  have  bx+ax=abm. 

Hence,  to  clear  an  equation  of  fractions,  we  have  the  follow 
ing 

RULE. 

Multiply  each  member  of  the  equation  by  all  the  denominators. 


EXAMPLES. 

x    x_S 
3~5~i 


x    x    3 

1.  Clear  the  equation  -— -=-  of  fractions. 

Ans.  20x- 12^=45. 

O™,         O™         O 

2.  Clear  the  equation  ————-=  of  fractions. 

Ans.  63*- 70^ =45. 


EQUATIONS   OF  THE   FIRST   DEGREE.  83 

3.  Clear  the  equation  -=- — —-\---6  of  fractions. 

i       4     o 

Ans.  40z-105x+28z=840. 

4.  Clear  the  equation  o+z+g  =  10  °f  fractions. 

138.  An  equation  may  always  be  cleared  of  fractions  by  muL 
tiplying  each  member  into  all  the  denominators;  but  some' 
times  the  same  result  may  be  attained  by  a  less  amount  of  mul 
tiplication.  Thus,  in  the  last  example,  the  equation  may  be 
cleared  of  fractions  by  multiplying  each  term  by  12  instead 
of  6x4x2,  and  it  is  important  to  avoid  all  useless  multiplica 
tion.  In  general,  an  equation  may  be  cleared  of  fractions  by 
multiplying  each  member  by  the  least  common  inultiple  of  all  the 
denominators. 

2x     Bx     7 

5.  Clear  the  equation  —  +—= —  of  fractions. 

The  least  common  multiple  of  all  the  denominators  is  20. 
If  we  multiply  each  member  of  the  equation  by  20,  we  obtain 


The  operation  is  effected  by  dividing  the  least  common  mul 
tiple  by  each  of  the  denominators,  and  then  multiplying  the 
corresponding  numerator,  dropping  the  denominator. 

6.  Clear  the  equation  y  — Tj— oT  °^  ^ract^ons- 

x— 4     1 

7.  Clear  the  equation  Bx— — -A — =^  °f  fractions. 

^  4       12 

It  should  be  remembered  that  when  a  fraction  has  the  mi 
nus  sign  before  it,  this  indicates  that  the  fraction  is  to  be  sub 
tracted,  and  the  signs  of  the  terms  derived  from  its  numerator 

must  be  changed,  Art.  118.  0    ,  ^  0     ., 

Ans.  36x— 3^+12  =  1. 

a— x     3x—2b     x-\-ab 

8.  Clear  the  equation  — = =- — = — - — . 

b  ab  a2 

Ans.  a3  —  azx— 


84  ALGEBRA. 

139.  Solution  of  Equations.  —  An  equation  of  the  first  degree 
containing  but  one  unknown  quantity  may  be  solved  by  trans 
forming  it  in  such  a  manner  that  the  unknown  quantity  shall 
stand  alone,  constituting  one  member  of  an  equation  ;  the  other 
member  will  then  denote  the  value  of  the  unknown  quantity, 
Let  it  be  required  to  find  the  value  of  x  in  the  equation 
4x—  2     5x_3x 
~~5~  ^""T4 
Clearing  of  fractions,  we  have 
32;c-16  +  25z 
By  transposition  we  obtain 


Uniting  similar  terms,    27#=216. 

Dividing  each  member  by  27,  according  to  Art  134,  we  have 

x=8. 

To  verify  this  value  of  x,  substitute  it  for  x  in  the  original 
equation,  and  we  shall  have 

32-2     40     24 


or 

that  is,  11  =  11, 

an  identical  equation,  which  proves  that  we  have  found  the 

correct  value  of  x. 

140.  Hence  we  deduce  the  following 

EULE. 

1.  Clear  the  equation  of  fractions,  and  perform  all  the  opera 
tions  indicated. 

2.  Transpose  all  the  terms  containing  the  unknown  quantity  to 
one  side,  and  all  the  remaining  terms  to  the  other  side  of  the  equa 
tion,  and  reduce  each  member  to  its  most  simple  form. 

3.  Divide  each  member  by  the  coefficient  of  the  unknown  quan 
tity. 

There  are  various  artifices  which   may  sometimes  be  em- 


EQUATIONS   OF   THE   FIRST   DEGREE.  85 

ployed,  by  which  the  labor  of  solving  an  equation  may  be  con 
siderably  abridged.  These  artifices  can  not  always  be  reduced 
to  general  rules.  If,  however,  any  reductions  can  be  made  be 
fore  clearing  of  fractions,  it  is  generally  best  to  make  them  ;  and 
if  the  equation  contains  several  denominators,  it  is  often  best  to 
multiply  by  the  simpler  denominators  first,  and  then  to  effect 
any  reductions  which  may  be  possible  before  getting  rid  of  the 
remaining  denominators.  Sometimes  considerable  labor  may 
be  saved  by  simply  indicating  a  multiplication  during  the  first 
steps  of  the  reduction,  as  we  can  thus  more  readily  detect  the 
presence  of  common  factors  (if  there  are  any),  which  may  be 
canceled.  The  discovery  of  these  artifices  will  prove  one  of 
the  most  useful  exercises  to  the  pupil. 

EXAMPLES. 

1.  Solve  the  equation 


_ 

~1~     ~T~ 
Clearing  of  fractions, 


Transposing  and  reducing, 

25x=150. 

Dividing  by  25,  x—6. 

To  verify  this  result,  put  6  in  the  place  of  x  in  the  original 
equation. 

Solve  the  following  equations: 

2.  Sax—  4:ab—2ax—  6ac.  Ans.  x=  46—  6c. 

.  Ans.  x=9. 


a(dz-\-x*)            ax  d 

4.  -A—  -  -  J-—ac-\--j-.  Ans.  x—-. 

dx                   d  c 

-    x_5  284-x 

5.  —  j  —  \-6x=  —  -  —  .  Ans.  x=9. 

a    ab     7    171  ab  —  1 

6.  —  =  bc+d+-.  Ans.  x= 


.  .         , 

x  x  fa  + 

Ans.  x~ 


ALGEBRA. 


0. 

9. 
10. 
11. 

13. 
14. 
15. 
16. 
17. 
is 

UU-JL  ^U-\-^±l>JL,        AA-\-UC. 

3x-5            2x-4 

T    \                                  I    ' 

£LiUS.    u;  —  -=.  —r:  ^, 

5a+46—  2 

t/O  '  f                             —  _L  4—                     o           * 

^j                                O 

8-r       11         E\r       E\       Q7        TV 
i/.  ~~-  J-  JL         t^tX/  1~^  c/         t7  i   ""™"  4  */>* 
•y  1  _l_                                     1 

^o  ^       4a2-35 

1  ^i                  R                   9        * 

Q       x—  4          5x+U     1 

4                  3      "12- 

a+x    5  —  x 

o           a 
3x          x             2x 

8a4-3ac—  3* 
a  free/ 

a           b~~          d' 

a2c 
(a-\-x)(b-\-x)  —  a(b-{-c)=  —--{-x2 

~"3&d+ad-4a&d-2a&' 
ac 

-d.715.  iC=T-« 

O^,  O                         OQ  q>          6^  8 

"~~s~"* 

7x+16      x4-8  _x 
~~21        4x-ll~~3' 
6x4-7     7x-13    2x4-4 

19.  § 


9 


(^  ,  fac— Icx=:fac+2a5— 6cx. 
6534 


J.ns.  05= 


7Qq5— 3ac 

320c     * 


22. 


T"+    3  4 


=  7. 
+^+1. 


6 


5  8 


11 


73:_6        x-5     =x 
35        6x— 101     5' 


8a 
Ans.  x=^. 


Ans.  x=5. 

Ans.  x=S%. 

Ans.  x=l±. 

Ans.  xr^ll. 


EQUATIONS  OF  THE   FIEST  DEGREE.  87 


27- 


xx          a  a2(b—a) 

28.  __u_  -  —  -  -  .  Ans.  x=-j-7j—t  —  r-. 
a^b-a    b+a  b(b+a) 

29.  x+x-4--(x+5)(x-3).  Ans.x=l2. 


25—  ^-x    16x+44      23 

-••30.-    TT-+  o  To  =~TT+5-  Ans.x=S$. 

x+1         3x+2       x+1 

V 

Solution  of  Problems. 

141.  A  problem  in  Algebra  is  a  question  proposed  requiring 
us  to  determine  the  value  of  one  or  more  unknown  quantities 
from  given  conditions. 

142.  The  solution  of  a  problem  is  the  process  of  finding  the 
value  of  the  unknown  quantity  or  quantities  that  will  satisfy 
the  given  conditions. 

143.  The  solution  of  a  problem  consists  of  two  parts  : 

1st.  The  statement,  which  consists  in  expressing  the  condi 
tions  of  the  problem  algebraically  ;  that  is,  in  translating  the 
conditions  of  the  problem  from  common  into  algebraic  lan 
guage,  or  forming  the  equation. 

2d.  The  solution  of  the  equation. 

The  second  operation  has  already  been  explained,  but  the 
first  is  often  more  embarrassing  to  beginners  than  the  second. 
Sometimes  the  conditions  of  a  problem  are  expressed  in  a  dis 
tinct  and  formal  manner,  and  sometimes  they  are  only  implied, 
or  are  left  to  be  inferred  from  other  conditions.  The  former 
are  called  explicit  conditions,  and  the  latter  implicit  conditions. 

144.  It  is  impossible  to  give  a  general  rule  which  will  enable 


88  ALGEBRA. 

us  to  translate  every  problem  into  algebraic  language,  since  the 
conditions  of  a  problem  may  be  varied  indefinitely.  The  fol 
lowing  directions  may  be  found  of  some  service  : 

Represent  one  of  the  unknown  quantities  l)y  some  letter  or  sym 
bol,  and  then  from  the  given  conditions  find  an  expression  for  each 
of  the  other  unknown  quantities,  if  any,  involved  in  the  problem. 

Express  in  algebraic  language  the  relations  which  subsist  between 
the  unknown  quantities  and  the  given  quantities  ;  or,  by  means  of 
the  algebraic  signs,  indicate  the  operations  necessary  to  verify  the 
value  of  the  unknown  quantity,  if  it  was  already  known. 

PROBLEMS. 

Prob.  1.  What  number  is  that,  to  the  double  of  which  if  16 
be  added,  the  sum  is  equal  to  four  times  the  required  number? 

Let  x  represent  the  number  required. 

The  double  of  this  will  be  2x. 

This  increased  by  16  should  equal  4x. 

Hence,  by  the  conditions,  2x+W=4:X. 

The  problem  is  now  translated  into  algebraic  language,  and 
it  only  remains  to  solve  the  equation  in  the  usual  way. 

Transposing,  we  obtain 

I6=4:x-2x=2x, 
and  8=x, 

or  x—S. 

To  verify  this  number,  we  have  but  to  double  8,  and  add 
16  to  the  result  ;  the  sum  is  32,  which  is  equal  to  four  times  8, 
according  to  the  conditions  of  the  problem. 

Prob.  2.  What  number  is  that,  the  double  of  which  exceeds 
its  half  by  6? 

Let  cc  =  the  number  required. 

Then,  by  the  conditions, 


Clearing  of  fractions,     4x—x=~L2, 
or  3*  =  12. 

Hence  cc=4. 

To  verify  this  result,  double  4,  which  makes  8,  and  diminish 


EQUATIONS   OF   THE   FIRST   DEGREE.  89 

it  by  the  half  of  4,  or  2  ;  the  result  is  6,  according  to  the  con 
ditions  of  the  problem. 

Prob.  3.  The  sum  of  two  numbers  is  8,  and  their  difference 
2.  What  are  those  numbers? 

Let  #=:the  least  number. 

Then  x+2  will  be  the  greater  number. 

The  sum  of  these  is  2os+2,  which  is  required  to  equal  8. 
Hence  we  have  2x4-2  =  8. 

By  transposition,  2x=8  — 2  — 6, 

and  x~B,  the  least  number. 

Also,  #+2=5,  the  greater  number. 

Verification.  5  +  3  =  8) 

K     Q     0:f  according  to  the  conditions. 

O  —  O  =  Z  ; 

The  following  is  a  generalization  of  the  preceding  Problem. 

Prob.  4.  The  sum  of  two  numbers  is  a,  and  their  difference 
I).  What  are  those  numbers? 

Let  x  represent  the  least  number. 

Then  x+b  will  represent  the  greater  number. 

The  sum  of  these  is  2x-\-b,  which  is  required  to  equal  a. 

Hence  we  have  2x+b=a. 

By  transposition,  ^x—a—b, 

a— b     a     b     ,      .  , 

or  x=— —  =  -  —  ft  the  less  number. 

lj          a      & 

Hence  x+b=^— -+&=-  +  -,  the  greater  number. 

£t      A  A      A 

As  these  results  are  independent  of  any  particular  value  at 
tributed  to  the  letters  a  and  b,  it  follows  that 

Half  the  difference  of  two  quantities,  added  to  half  their  sum, 
is  equal  to  the  greater ;  and 

Half  the  difference  subtracted  from  half  the  sum  is  equal  to  the 
less. 

The  expressions  |+|  and  §— ~  are  culled  formulas,  because 

LA  A      A 

they  may  be  regarded  as  comprehending  the  solution  of  all 
questions  of  the  same  kind;  that  is,  of  all  problems  in  which 
we  have  given  the  sum  and  difference  of  two  quantities. 


90 


ALGEBRA. 


Thus,  let  a=8  \       . 

1—2  \  as  m  tne  Precedmg  problem. 

Then  o^~2~~~2~~^'  ^e  Sreater  number. 

a    &     8—2     0  ,,     ,  , 

And  -—-=—^—  =  3,  the  less  number. 


10; 

12; 

23; 

100 ;   their  difference  - 
100; 
5; 

10; 


6; 

2; 
ii; 

50  ;   required  the  numbers. 

1; 
*; 
ii 


Prob.  5.  From  two  towns  which  are  54  miles  distant,  two 
travelers  set  out  at  the  same  time  with  an  intention  of  meet 
ing.  One  of  them  goes  4  miles  and  the  other  5  miles  per  hour. 
In  how  many  hours  will  they  meet? 

Let  x  represent  the  required  number  of  hours. 

Then  4#  will  represent  the  number  of  miles  one  traveled, 
and  5x  the  number  the  other  traveled ;  and  since  they  meet, 
they  must  together  have  traveled  the  whole  distance. 

Consequently,  4x+ 5#— 54. 

Hence  9x=54, 

or  a;=6. 

.  Proof.  In  6  hours,  at  4  miles  an  hour,  one  would  travel  24 
miles;  the  other,  at  5  miles  an  hour,  would  travel  30  miles. 
The  sum  of  24  and  30  is  54  miles,  which  is  the  whole  distance. 

This  Problem  may  be  generalized  as  follows : 

Prob.  6.  From  two  points  which  are  a  miles  apart,  two  bod 
ies  move  toward  each  other,  the  one  at  the  rate  of  m  miles  per 
hour,  the  other  at  the  rate  of  n  miles  per  hour.  In  how  many 
hours  will  they  meet? 

Let  x  represent  the  required  number  of  hours. 

Then  mx  will  represent  the  number  of  miles  one  body  moves, 
and  nx  the  miles  the  other  body  moves,  and  we  shall  obvious 
ly  have  mx-\-nx—a. 


EQUATIONS  OF  THE   FIRST  DEGREE.  9} 


Hence  x— 

This  is  a  general  formula,  comprehending  the  solution  of  all 
problems  of  this  kind.  Thus, 

(150; 
Let  the      J   90;   onebody 

15; 
210;  20; 

Eequired  the  time  of  meeting. 

We  see  that  an  infinite  number  of  problems  may  be  pro 
posed,  all  similar  to  Prob.  5 ;  but  they  are  all  solved  by  the 
formula  of  Prob.  6.  We  also  see  what  is  necessary  in  order 
that  the  answers  may  be  obtained  in  whole  numbers.  The  given 
distance  (a)  must  be  exactly  divisible  by  m-{-n. 

Prob.  7.  A  gentleman,  meeting  three  poor  persons,  divided 
60  cents  among  them  ;  to  the  second  he  gave  twice,  and  to  the 
third  three  times  as  much  as  to  the  first.  What  did  he  give 
to  each? 

Let  £e=the  sum  given  to  the  first;  then  2x=ihe  sum  given 
to  the  second,  and  3x— the  sum  given  to  the  third. 

Then,  by  the  conditions, 

x-f  2cc+3x=60. 

That  is,  6x=60, 

or  05=10. 

Therefore  he  gave  10,  20,  and  30  cents  to  them  respectively. 
The  learner  should  verify  this,  and  all  the  subsequent  results. 

The  same  problem  generalized : 

Prob.  8.  Divide  the  number  a  into  three  such  parts  that  the 
second  may  be  m  times,  and  the  third  n  times  as  great  as  the 

first. 

A  a  ma  na 

Ans.  -r ;  : ;  . 

L -\-7n-\-n  L-\-m-\-n      \-\-m-\-n 

What  is  necessary  in  order  that  the  preceding  values  may 
be  expressed  in  whole  numbers? 

Prob.  9.  A  bookseller  sold  10  books  at  a  certain  price,  and 
afterward  15  more  at  the  same  rate.  Now  at  the  last  sale  he 


92  ALGEBRA. 

received  25  dollars  more  than  at  the  first.  What  did  he  re 
ceive  for  each  book?  •*  Ans.  Five  dollars. 

The  same  Problem  generalized: 

Prob.  10.  Find  a  number  such  that  when  multiplied  success 
ively  by  m  and  by  n,  the  difference  of  the  products  shall  be  a. 

A  a 

Ans.  . 

m — n 

Prob.  11.  A  gentleman,  dying,  bequeathed  1000  dollars  to 
three  servants.  A  was  to  have  twice  as  much  as  B,  and  B 
three  times  as  much  as  C.  What  were  their  respective  shares? 

Ans.  A  received  $600,  B  $300,  and  C  $100. 
Prob.  12.  Divide  the  number  a  into  three  such  parts  that  the 
second  may  be  m  times  as  great  as  the  first,  and  the  third  n 
times  as  great  as  the  second. 

.  a          ^          ma  mna 

l  +  m+mn'    \-\-m-\-mn'1    \-\-m  +  mri 

Prob.  13.  A  hogshead  which  held  120  gallons  was  filled  with 
a  mixture  of  brandy,  wine,  and  water.     There  were  10  gallons 
of  wine  more  than  there  were  of  brandy,  and  as  much  water  as 
both  wine  and  brandy.     What  quantity  was  there  of  each? 
Ans.  Brandy  25  gallons,  wine  35,  and  water  60  gallons. 
Prob.  14.  Divide  the  number  a  into  three  such  parts  that  the 
second  shall  exceed  the  first  by  m,  and  the  third  shall  be  equal 
to  the  sum  of  the  first  and  second. 

a — 2m     a-f-2m     a 
ns.  —J—\    — |—  ;    2' 

Prob.  15.  A  person  employed  four  workmen,  to  the  first  of 
whom  he  gave  2  shillings  more  than  to  the  second ;  to  the  sec 
ond  3  shillings  more  than  to  the  third  ;  and  to  the  third  4  shil 
lings  more  than  to  the  fourth.  Their  wages  amount  to  32  shil 
lings.  What  did  each  receive  ? 

Ans.  They  received  12, 10,  7,  and  3  shillings  respectively. 

Prob.  16.  Divide  the  number  a  into  four  such  parts  that  the 
second  shall  exceed  the  first  by  m,  the  third  shall  exceed  the 
second  by  n,  and  the  fourth  shall  exceed  the  third  by  p. 

a  —  3m  —  2n—p       .  , 

Ans.  The  first, z L  ;    the  second, 


EQUATIONS   OF  THE   FIRST  DEGREE.  93 


^    Ai_-  j  —         ,     f       , 

the  third,  —  '  -  —      -  ;    the  fourth, 


Problems  which  involve  several  unknown  quantities  may  oft 
en  be  solved  by  the  use  of  a  single  unknown  letter.  Most  of 
the  preceding  examples  are  of  this  kind.  In  general,  when  we 
have  given  the  sum  or  difference  of  two  quantities,  both  of  them 
may  be  expressed  by  means  of  the  same  letter.  For  the  differ* 
ence  of  two  quantities  added  to  the  less  must  be  equal  vo  the 
greater  ;  and  if  one  of  two  quantities  be  subtracted  from  their 
sum,  the  remainder  will  be  equal  to  the  other. 

Prob.  17.  At  a  certain  election  36,000  votes  were  polled, 
and  the  candidate  chosen  wanted  but  3000  of  having  twice  as 
many  votes  as  his  opponent.  How  many  voted  for  each  ? 

Let  x=thevnumber  of  votes  for  the  unsuccessful  candidate; 
then  36,000  —  x  =  the  number  the  successful  one  had,  and 
36,000—  x+  3000  =  2x.  Ans.  13,000  and  23,000. 

Prob.  18.  Divide  the  number  a  into  two  such  parts  that  one 
part  increased  by  b  shall  be  equal  to  m  times  the  other  part. 

.       ma  —  b      a  +  b 
Ans.  -  —  r  ;   -  -. 

772  +  1    '     m  +  1 

Prob.  19.  A  train  of  cars,  moving  at  the  rate  of  20  miles  per 
hour,  had  been  gone  3  hours,  when  a  second  train  followed  at 
the  rate  of  25  miles  per  hour.  In  what  time  will  the  second 
train  overtake  the  first? 

Let  x=the  number  of  hours  the  second  train  is  in  motion, 
and  05  -f  3  =  the  time  of  the  first  train. 

Then  25x=:the  number  of  miles  traveled  by  the  second  train, 
and  20(cc  +  3)  =  the  miles  traveled  by  the  first  train. 

But  at  the  time  of  meeting  they  must  both  have  traveled  the 
same  distance. 

Therefore  25x  =  20x  -f  60. 

By  transposition,  5x=60, 

and  05=12. 

Proof.  In  12  hours,  at  25  miles  per  hour,  the  second  train 
goes  300  miles  ;  and  in  15  hours,  at  20  miles  per  hour,  the  first 
train  also  goes  300  miles;  that  is,  it  is  overtaken  by  the  sec 
ond  train. 


94  ALGEBRA. 

Prob.  20.  Two  bodies  move  in  the  same  direction  from  two 
places  at  a  distance  of  a  miles  apart  ;  the  one  at  the  rate  of  n 
miles  per  hour,  the  other  pursuing  at  the  rate  of  ra  miles  per 
hour.  When  will  they  meet  ? 

Ans.  In  —  J—  hours. 
m  —  n 

This  Problem,  it  will  be  seen,  is  essentially  the  same  as 
Prob.  10. 

Prob.  21.  Divide  the  number  197  into  two  such  parts  that 
four  times  the  greater  may  exceed  five  times  the  less  by  50. 

Ans.  82  and  115. 

Prob.  22.  Divide  the  number  a  into  two  such  parts  that  m 
times  the  greater  may  exceed  n  times  the  less  by  b. 

ma  —  b     na+b 

Ans.  -  ;   -  . 

m+n      m-\-n 

"When  w  =  l,  this  Problem  reduces  to  Problem  18. 
When  5  —  0,  this  Problem  reduces  to  Problem  24. 

Prob.  23.  A  prize  of  2329  dollars  was  divided  between  two 
persons,  A  and  B,  whose  shares  were  in  the  ratio  of  5  to  12. 
What  was  the  share  of  each  ? 

Beginners  almost  invariably  put  x  to  represent  one  of  the 
quantities  sought  in  a  problem;  but  a  solution  may  often  be 
very  much  simplified  by  pursuing  a  different  method.  Thus, 
in  the  preceding  problem,  we  may  put  x  to  represent  one  fifth 
of  A's  share.  Then  5x  will  be  A's  share,  and  I2x  will  be  B's, 
and  we  shall  have  the  equation 


and  hence  x  =137; 

consequently  their  shares  were  685  and  1644  dollars. 

Prob.  24.  Divide  the  number  a  into  two  such  parts  that  the 
first  part  may  be  to  the  second  as  m  to  n. 


ma          na 
Ans. 


m-{-n 
Prob.  25.  What  number  is  that  whose  third  part  exceeds  its 

fourth  part  by  16  ? 
Let  12x=the  number. 

Then  4x-3x=l6. 


EQUATIONS  OF  THE   FIKST  DEGREE.  95 

or  x=I6. 

Therefore  the  number  -  12  x  16-192.  ^ 

.   Prob.  26.  Find  a  number  such  that  when  it  is  divided  suc 

cessively  by  m  and  by  n,  the  difference  of  the  quotients  shall 

be  a. 

amn 
Ans. 


n  — 


Prob.  27.  A  gentleman  has  just  8  hours  at  his  disposal  ;  how 
far  may  he  ride  in  a  coach  which  travels  9  miles  an  hour,  so  as 
to  return  home  in  time,  walking  back  at  the  rate  of  3  miles  an 
hour?  Ans.  18  miles. 

w  Prob.  28.  A  gentleman  has  just  a  hours  at  his  disposal  ;  how 
far  may  he  ride  in  a  coach  which  travels  m  miles  an  hour,  so 
as  to  return  home  in  time,  walking  back  at  the  rate  of  n  miles 

an  hour? 

amn       ., 
Ans.  —   -  miles. 
m+n 

-  Prob.  29.  A  gentleman  divides  a  dollar  among  12  children, 
giving  to  some  9  cents  each,  and  to  the  rest  7  cents.  How 
many  were  there  of  each  class? 

-Prob.  30.  Divide  the  number  a  into  two  such  parts  that  if 
the  first  is  multiplied  by  m  and  the  second  by  n,  the  sum  of 

the  products  shall  be  b. 

o  —  na     ma  —  o 
Ans.  -  :   -  . 


m  —  n      m  —  n 


-Prob.  31.  If  the  sun  moves  every  day  1  degree,  and  the 
moon  13,  and  the  sun  is  now  60  degrees  in  advance  of  the 
moon,  when  will  they  be  in  conjunction  for  the  first  time,  sec 
ond  time,  and  so  on  ? 

-  Prob.  32.  If  two  bodies  move  in  the  same  direction  upon  the 
circumference  of  a  circle  which  measures  a  miles,  the  one  at 
the  rate  of  n  miles  per  day,  the  other  pursuing  at  the  rate  of  m 
miles  per  day,  when  will  they  be  together  for  the  first  time,  sec 
ond  time,  etc.,  supposing  them  to  be  b  miles  apart  at  starting? 

AM.  In  ~>     £±*,  2*t*,  etc,  days. 

m  —  Ji  in  —  n  m—n 

It  will  be  seen  that  this  Problem  includes  Prob.  20. 


96  ALGEBRA. 

~  Prob.  33.  Divide  the  number  12  into  two  such  parts  that  the 
difference  of  their  squares  may  be  48. 

*  Prob.  34.  Divide  the  number  a  into  two  such  parts  that  the 

difference  of  their  squares  rnay.be  b.  a2— b     a2  +  b 

J  i,  Ans.  — —  ;    -- — . 

2a          2a 

Prob.  35.  The  estate  of  a  bankrupt,  valued  at  21,000  dollars, 
is  to  be  divided  among  three  creditors  according  to  their  re 
spective  claims.  The  debts  due  to  A  and  B  are  as  2  to  3, 
while  B's  claims  and  C's  are  in  the  ratio  of  4  to  5.  What  sum 
must  each  receive? 

-  Prob.  36.  Divide  the  number  a  into  three  parts,  which  shall 
be  to  each  other  as  m :  n:  p. 

ma  na  pa 

Ans. 


When  _p  =  l,  Prob.  36  reduces  to  the  same  form  as  Prob.  8. 
^  Prob.  37.  A  grocer  has  two  kinds  of  tea,  one  worth  72  cents 
per  pound,  the  other  40  cents.  How  many  pounds  of  each 
must  be  taken  to  form  a  chest  of  80  pounds,  which  shall  be 
worth  60  cents? 

Ans.  50  pounds  at  72  cents,  and  30  pounds  at  40  cents. 

Prob.  38.  A  grocer  has  two  kinds  of  tea,  one  worth  a  cents 
per  pound,  the  other  b  cents.  How  many  pounds  of  each  must 
be  taken  to  form  a  mixture  of  n  pounds,  which  shall  be  worth 

c  cents?  n(c—b) 

Ans.  —  -  —  j-Z  pounds  at  a  cents, 

n(a  —  c)  , 

and  —  -  —  —  ^  pounds  at  b  cents. 
a  —  b 

-Prob.  39.  A  can  perform  a  piece  of  work  in  6  daj^s;  B  can 
perform  the  same  work  in  8  days;  and  C  can  perform  the 
same  work  in  24  days.  In  what  time  will  they  finish  it  if  all 
work  together? 

~Prob.  40.  A  can  perform  a  piece  of  work  in  a  days,  B  in  b 
days,  and  C  in  c  days.  In  what  time  will  they  perform  it  if  all 
work  together?  ^  «fe  _  d 


Prob.  41.  There  are  three  workmen,  A,  B,  and  C.     A  and 
B  together  can  perform  a  piece  of  work  in  27  days;  A  and  C 


EQUATIONS  OF  THE   FIRST  DEGREE.  97 

together  in  86  days;  and  B  and  C  together  in  54  days.     In 
what  time  could  they  finish  it  if  all  worked  together  ? 

A  and  B  together  can  perform  -5-7-  of  the  work  in  one  day. 

AandC  "  A 

B  and  C  *     ^ 

Therefore,  adding  these  three  results, 

2A  +  2B  +  2C  can  perform  -^V  +  ^V  S-^V  iri  one  day, 
=1-^  in  one  day. 

Therefore,  A,  B,  and  C  together  can  perform  -^  of  the  work 
in  one  day  ;  that  is,  they  can  finish  it  in  24  days.  If  we  put 
x  to  represent  the  time  in  which  they  would  all  finish  it,  then 
they  would  together  perform  |  part  of  the  work  in  one  day, 
And  we  should  have 


Prob.  42.  A  and  B  can  perform  a  piece  of  labor  in  a  days  ; 
A  and  C  together  in  b  days;  and  B  and  C  together  in  c  days. 
In  what  time  could  they  finish  it  if  all  work  together  ? 


A  -, 

Am.  -T—      —  y-  days. 


This  result,  it  will  be  seen,  is  of  the  same  form  as  that  of 
Problem  40.  ^ 

Prob.  43.  A  broker  has  two  kinds  of  change.  It  takes  20 
pieces  of  the  first  to  make  a  dollar,  and  4  pieces  of  the  second 
to  make  the  same.  Now  a  person  wishes  to  have  8  pieces  for 
a  dollar.  How  many  of.  each  kind  must  the  broker  give  him? 

Prob.  44.  A  has  two  kinds  of  change  ;  there  must  be  a  pieces 
of  the  first  to  make  a  dollar,  and  b  pieces  of  the  second  to  make 
the  same.  Now  B  wishes  to  have  c  pieces  for  a  dollar.  How 
many  pieces  of  each  kind  must  A  give  him  ? 

^.725.  —  —  7—  of  the  first  kind  ;   ~  —  ^  of  the  second. 
a  —  b  a  —  b 

Prob.  45.  Divide  the  number  45  into  four  such  p^rts  that 
the  first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  all  be  equal. 

In  solving  examples  of  this  kind,  several  unknown  quantities 
are  usually  introduced,  but  this  practice  is  worse  than  super- 

E 


98  ALGEBRA. 

fluous.  The  four  parts  into  which  45  is  to  be  divided  may  be 
represented  thus: 

The  first  =a-2, 

the  second  =cc-f  2, 

the  third  =f, 

the  fourth  =  2x; 

for  if  the  first  expression  be  increased  by  2,  the  second  dimin 
ished  by  2,  the  third  multiplied  by  2,  and  the  fourth  divided  by 
2,  the  result  in  each  case  will  be  x.  The  sum  of  the  four  parts 
is  4^x,  which  must  equal  45. 

Hence  x=10. 

Therefore  the  parts  are  8,  12,  5,  and  20. 

Prob.  46.  Divide  the  number  a  into  four  such  parts  that 
the  first  increased  by  m,  the  second  diminished  by  m,  the  third 
multiplied  by  m,  and  the  fourth  divided  by  m,  shall  all  be 

equal. 

ma  ma  a  m2a 

Ans' 


(m  +  1)2         ;   (ra  +  1)2^      ;   (m  +  1)2'  (m+lf 

Prob.  47.  A  merchant  maintained  himself  for  three  years  at 
an  expense  of  $500  a  year,  and  each  year  augmented  that 
part  of  his  stock  which  was  not  thus  expended  by  one  third 
thereof.  At  the  end  of  the  third  year  his  original  stock  was 
doubled.  What  was  that  stock? 

Prob.  48.  A  merchant  supported  himself  for  three  years  at 
an  expense  of  a  dollars  per  year,  and  each  year  augmented 
that  part  of  his  stock  which  was  not  thus  expended  by  one 
third  thereof.  At  the  end  of  the  third  year  his  original  stock 

was  doubled.     What  was  that  stock  ? 

USa 
Ans.  — . 

Prob.  49.  A  father,  aged  54  years,  has  a  son  aged  9  years. 
In  how  many  years  will  the  age  of  the  father  be  four  times 
that  of  the  son  ? 

Prob.  50.  The  age  of  a  father  is  represented  by  a,  the  age  of 
his  son  by  b.  In  how  many  years  will  the  age  of  the  father  be 

n  times  that  of  the  son  ?  a— nb 

Ans.  =-. 

n  —  1 


EQUATIONS  WITH  MOKE  THAN  ONE  UNKNOWN  QUANTITY.  99 


CHAPTER  IX. 

AQUATIONS    OF    THE    FIRST   DEGREE    CONTAINING  MORE   THAN 
ONE    UNKNOWN    QUANTITY. 

145.  If  we  have  a  single  equation  containing  two  unknown 
quantities,  then  for  every  value  which  we  please  to  ascribe  to 
one  of  the  unknown  quantities,  we  can  determine  the  corre 
sponding  value  of  the  other,  and  thus  find  as  many  pairs  of 
values  as  wevplease  which  will  satisfy  the  equation.  Thus,  let 

2a+4y=16.  (1.) 

If  y—\,  we  find  cr=6;  if  y=2,  we  find  x=4,  and  so  on; 
and  each  of  these  pairs  of  values,  1  and  6,  2  and  4,  etc.,  sub 
stituted  in  equation  (1),  will  satisfy  it. 

Suppose  that  we  have  another  equation  of  the  same  kind, 
as,  for  example,  5x+3y=19.  (2.) 

We  can  also  find  as  many  pairs  of  values  as  we  please  which 
will  satisfy  this  equation. 

But  suppose  we  are  required  to  satisfy  both  equations  with 
the  same  set  of  values  for  .a;  and  y ;  we  shall  find  that  there  is 
only  one  value  of  x  and  one  value  of  y.  For,  multiply  equa 
tion  (1)  by  3,  and  equation  (2)  by  4,  Axiom  3,  and  we  have 

6x+12?/=48,  (3.) 

20x+12y=76.  (4.) 

Subtracting  equation  (3)  from  equation  (4),  Axiom  2,  we  have 

14^=28;  (5.) 

whence  x=2.  (6.) 

Substituting  this  value  of  x  in  equation  (1),  we  have 

4+4y=16;  (7.) 

Whence  y=3.  (8.) 

Thus  we  see  that  if  both  equations  are  to  be  satisfied,  x  must 
equal  2,  and  y  must  equal  3<  Equations  thus  related  are  called 
simultaneous  equations 


100  ALGEBRA. 

146.  Simultaneous  equations  are  those  which  must  be  satisfied 
by  the  same  values  of  the  unknown  quantities. 

When  two  or  more  simultaneous  equations  are  given  for  so 
lution,  we  must  endeavor  to  deduce  from  them  a  single  equation 
containing  only  one  unknown  quantity.  We  must  therefore 
make  one  of  the  unknown  quantities  disappear,  or,  as  it  is 
termed,  we  must  eliminate  it. 

147.  Elimination  is  the  operation  of  combining  two  or  more 
equations  in  such  a  manner  as  to  cause  one  of  the  unknown 
quantities  contained  in  them  to  disappear. 

There  are  three  principal  methods  of  elimination  :  1st,  by  ad 
dition  or  subtraction  ;  2d,  by  substitution  ;  3d,  by  comparison. 

148.  Elimination  ly  Addition  or  Subtraction. — Let  it  be  pro 
posed  to  solve  the  system  of  equations 

5x+%=35,  (1.) 

7x-3y=6.  (2.) 

Multiplying  equation  (1)  by  3,  and  equation  (2)  by  4,  we  have 

(3.) 
(4.) 
Adding  (3)  and  (4),  member  to  member  (Axiom  1),  we  have 

43^=129;  (5.) 

whence  <r  — 3.  (6.) 

We  may  now  deduce  the  value  of  y  by  substituting  the  value 
of  x  in  one  of  the  original  equations.     Taking  the  first  for  ex 
ample,  we  have  15+4^=35; 
whence  4y=20, 
and  y=§> 

149.  In  the  same  wa}^,  an  unknown  quantity  may  be  elimi 
nated  from  any  two  simultaneous  equations.     This  method  is 
expressed  in  the  following 

RULE. 

Multiply  or  divide  the  equations,  if  necessary,  in  such  a  manner 
iliat  one  of  the  unknoivn  quantities  sliatt  have  the  same  coefficient  in 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.  101 

both.  Then  subtract  one  equation  from  the  other  if  the  signs  of 
these  coefficients  are  alike,  or  add  them  together  if  tJie  signs  art 
unlike. 

In  solving  the  preceding  equations,  we  multiplied  both  mem 
bers  of  each  by  the  coefficient  of  the  quantity  to  be  eliminated 
in  the  other  equation  ;  but  if  the  coefficients  of  the  letter  to  be 
eliminated  have  any  common  factor,  we  may  accomplish  the 
same  object  by  the  use  of  smaller  multipliers.  In  such  cases, 
find  the  least  common  multiple  of  the  coefficients  of  the  letter 
to  be  eliminated,  and  divide  this  multiple  by  each  coefficient; 
the  quotients  will  be  the  least  multipliers  which  we  can  employ. 

150.  Elimination  by  Substitution.  —  Take  the  same  equations 
as  before  :  6x+4y=  35,  (1.) 

7a-3y=6.  (2.) 

Finding  from  (1)  the  value  of  y  in  terms  of  x,  we  have 

y=^.  (3.) 


Substituting  this  value  of  y  in  (2),  we  have 
105-15$ 

7x  —  rTrtv>>  • 

Clearing  of  fractions, 


whence  x=3. 

Substituting  this  value  of  x  in  (3),  we  have 

y=6. 

The  method  thus  exemplified  is  expressed  in  the  following 

RULE. 

find  an  expression  for  the  value  of  one  of  the  unknown  quan 
tities  in  one  of  the  equations;  then  substitute  this  value  for  that 
quantity  in  the  other  equation. 

151.  Elimination  by  Comparison.  —  Take  the  same  equations 
as  before:  5z+4?/=35,  (1.) 

7x-3y=6.  (2.) 


102  ALGEBRA. 

Derive  from  each  equation  an  expression  for  y  in  terms  of  a\ 
and  we  have  35—  5x 

y=  -—  , 


- 
and  y=—  g—  .  (4.) 

Placing  these  two  values  equal  to  each  other,  we  have 


3  4 

Clearing  of  fractions, 

28x-24=:105-15z; 
whence  43x-129, 

and  x=S. 

Substituting  this  value  of  x  in  (3), 

y=5. 

The  method  thus  exemplified  is  expressed  in  the  following 
4- 

KULE. 

Find  an  expression  for  the  value  of  the  same  unknown  quantity 
in  each  of  the  equations,  and  form  a  new  equation  by  placing  these, 
values  egwal  to  edch  ether. 

In  the  solution  of  simultaneous  equations,  either  of  the  pre 
ceding  methods  can  be  u&eri,  as  may  be  most  convenient,  and 
each  method  has  its  advantages  in  particular  cases.  General 
ly,  however,  the  last  two  methods  give  rise  to  fractional  expres 
sions,  which  occasion  inconvenience  in  practice,  while  the  first 
method  is  not  liable  to  this  objection.  When  the. coefficient  of 
one  of  the  unknown  quantities  in  one  of  the  equations  is  equal 
to  unity,  this  inconvenience  does  not  occur,  and  the  method 
'Jby  substitution  may  be  preferable ;  the  first  will,  however,  com- 
irnonly  be  found  most  convenient. 

EXAMPLES. 

f  llsr+3?/=100)  x 
1.  Given  j     4   _</  _4     f  to  ^d  the  values  of  x  and  y. 

Ans.  x=8:  ?/=4. 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.  103 


2.  Given 


3.  Given 


4.  Given 


1+!-' 
H=8 

x  +  2 


y+5 

4 


to  find  x  and  y. 


Am.  x  —  6;  y=12. 
to  find  x  and  y. 

to  find  x  and  y. 


a     o 

5.  Given  <  *    ^ 
c     d 

--] —  —  n 
x    y 


3  2 

i 

to  find  x  and  y. 


be— ad  be— ad 

Am.  x=—, ,;   y= 


nb — mo?'          me — na 

j   5x-7y=20  )  4    G    , 
6.  Given  -j  n       -  r       . .  [•  to  find  x  and  £/. 
(  9x—  ll?/=44  ) 


7.  Given 


-=m 

/y>  rt  y 

8.  Given  <J  V  to  find  x  and  ?/. 

-=i-»  2  2 

y    x         J  Am.x=-    -;   y= 


m  —  n 


T^^o.    n 

9.  Given  <J      ^~  \  to  find  x  and  y. 

\LX-\-v(         . 

f~    i~i-r     '  i  >4 775  x=:21  *   v — 3^ 


10.  Given 


to  find  x  and 


Ans.  vc=na—a;  y  = 


104 

11.  Given 

12.  Given 


ALGEBRA. 


to  find  x  and    - 


11— x    4x+8y—2 
~~2~         ~~9~ 


13.  Given 


6  3 


to  find  x  and 

™  —  1  •    ?/  — 9 

<L — i,  y—&. 
to  find  x  and  y. 


14.  Given 


13 


— 5?/+6 
19 


6x— oy+4    3x+2?/+l 


to  find  cc  and  T. 


15.  Given  \  ?    [  to  find  x  and  y. 

(  x—y—l    \ 


Ans.  x=—  ;   y= 


a-b* 


16.  Given 


to  find  x  and    . 


17.  Given 


5         ~10156 
a;_8y_1_y-a?    a;      1 
2     20  15  ^6^1 


to  find 
and    . 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.  105 

Equations  of  the  First  Degree  containing  more  than  Two 
Unknown  Quantities. 

152.  If  we  have  three  simultaneous  equations  containing 
three  unknown  quantities,  we  may,  by  the  preceding  methods, 
reduce  two  of  the  equations  to  one  containing  only  two  of  the 
jnknown  quantities;  then  reduce  the  third  equation  and  either 
of  the  former  two  to  one  containing  the  same  two  unknown 
quantities  ;  and  from  the  two  equations  thus  obtained,  the  un 
known  quantities  which  they  involve  may  be  found.  The 
third  quantity  may  then  be  found  by  substituting  these  values 
in  either  of  the  proposed  equations. 

Take  the  system  of  equations 


3x+2?/-5z=  8,  (2.) 

§x-§y  4-  3z  rr  6.  (3.) 

Multiplying  (1)  by  3,  and  (2)  by  2,  we  have 

6x+9z/-fl2z=48,  (4.) 

6x4-4^-102=16.  (5.) 

Subtracting  (5)  from  (4),      5z/+222  =  32.  (6.) 

Multiplying  (1)  by  5,  and  (3)  by  2,  we  have 

"lO;c+15y  4-202=80,  (7.) 

10:c-12y4-6*  =  12.  (8.) 

Subtracting  (8)  from  (7),      27y  4-142  =  68.  (9.) 

Multiplying  (6)  by  27,  and  (9)  by  5,  we  have 

135  ?/+  594s  =  864.  (10.) 

135?/+702  =  340.  (11.) 

Subtracting  (11)  from  (10),     5242=524  ; 
whence  z  =  l. 

Substituting  this  value  of  z  in  (6), 
5y4-22=32; 
whence  2/=2. 

Substituting  the  values  of  y  and  z  in  (1), 

2^4-64-4=16; 

whence  <c=3. 

E  2 


106  ALGEBRA. 

153.  Hence,  to  solve  three  equations  containing  three  un 
known  quantities,  we  have  the  following 

RULE. 

From  the  three  equations  deduce  two  containing  only  two  un* 
known  quantities  ;  then  from  these  two  deduce  one  containing  only 
one  unknown  quantity. 

154.  If  we  had/owr  simultaneous  equations  containing  four 
unknown  quantities,  we  might,  by  the  methods  already  ex 
plained,  eliminate  one  of  the  unknown  quantities.     We  should 
thus  obtain  three  equations  between  three  unknown  quanti 
ties,  which  might  be  solved  according  to  Art.  152.      So,  also, 
if  we  \\2idifive  equations  containing  five  unknown  quantities, 
we  might,  by  the  same  process,  reduce  them  to  four  equations 
containing  four  unknown  quantities,  then  to  three,  and  so  on. 
By  following  the  same  method,  we  might  resolve  a  system  of 
any  number  of  equations  of  the  first  degree.    Hence,  if  we  have 
m  equations  containing  m  unknown  quantities,  we  proceed  by 
the  following 

RULE. 

1st.  Combine  successively  any  one  of  the  equations  with  each  of 
Hie  others,  so  as  to  eliminate  the  same  unknown  quantity;  there 
will  result  m  —  \  new  equations,  containing  m — 1  unknown  quan 
tities. 

2d.  Combine  any  one  of  these  new  equations  with  the  others,  so  as 
to  eliminate  a  second  unknown  quantity  ;  there  will  result  m  —  2 
equations,  containing  m — 2  unknown  quantities. 

3d.  Continue  this  series  of  operations  until  there  results  a  single 
equation  containing  but  one  unknown  quantity,  from  which  the 
valne  of  this  unknown  quantity  is  easily  deduced. 

4th.  /Substitute  this  value  for  its  equal  in  one  of  the  equations  con  • 
taining  two  unknown  quantities,  and  thus  find  the  value  of  a  second 
unknown  quantity]  substitute  these  values  in  an  equation  contain 
ing  three  unknown  quantities,  and  find  the  value  of  a  third'  and 
so  on,  till  the  values  of  all  are  determined. 

Either  of  the  unknown  quantities  may  be  selected  as  the  one 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.  107 


to  be  first  eliminated.  It  is,  however,  generally  best  to  begin 
with  that  which  has  the  smallest  coefficients  ;  and  if  each  of 
the  unknown  quantities  is  not  contained  in  all  the  proposed 
equations,  it  is  generally  best  to  begin  with  that  which  is  found 
in  the  least  number  of  equations.  Sometimes  a  solution  may 
be  very  much  abridged  by  the  use  of  peculiar  artifices,  for 
which  no  general  rules  can  be  given. 

EXAMPLES. 

Solve  the  following  groups  of  simultaneous  equations  : 


Ans.       = 


2.      x+z=b 


Note.  Take  the  sum  of  the  three  preceding  equations. 

[       x+y+z=29\  (x=  8. 

3.   <   x+2y+3z  =  62V  Ans.  |yi  9- 

+z=10)  (*=12. 


5.  1     x-?j+z=  654V 

(  -X  +  7/  +  2=-12) 


Ans.        = 


7.  J 


0  7 


108 


ALGEBRA. 


8.  - 


9.  - 


M=« 

x     y 

-4--  =  b 

x     z~ 

l-+l-=c 


1  l_l=a 
a   y  z~ 

i_i+u 

x    y    z 


12 


=  1 


30      t      37     =3 


222 


10.  4 


p2cc+5y-7z=-288 
11.  i     5x-y+Zz  =     227 

:s=      297 


12.  J 


x    y     Iz 
3  +  5+Y 


2,^8^5' 


13. 


4y—3u+2v=  9 


Ans.  - 


Ans.  -< 


x= 


y— 


a+ti 

2 
a-f-c* 

2 


r*=l. 

J.n.'j.  -!  y  =  2. 

U=3. 


x-13. 
y=24. 


x=  12. 
y=  30. 


=  50. 


z  =  S. 
M=3. 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.    10§ 

f  x  =  3. 


1  x+y+t+u+v  = 


14. 


Note.  Take  the  sum  of  these  six  equations. 


Ans. 


v=8. 


Problems  involving  Equations  of  the  First  Degree  with  several 
Unknown  Quantities. 

Prob.  1.  Find  two  numbers  such  that  if  the  first  be  added 
to  four  times  the  second,  the  sum  is  29 ;  and  if  the  second  be 
added  to  six  times  the  first,  the  sum  is  36. 

Prob.  2.  If  A's  money  were  increased  by  36  shillings,  he 
would  have  three  times  as  much  as  B ;  but  if  B's  money  were 
diminished  by  5  shillings,  he  would  have  half  as  much  as  A. 
Find  the  sum  possessed  by  each. 

Prob.  3.  A  pound  of  tea  and  three  pounds  of  sugar  cost  six 
shillings ;  but  if  sugar  were  to  rise  50  per  cent,  and  tea  10  per 
cent.,  they  would  cost  seven  shillings.  Find  the  price  of  tea 
and  sugar.  Ans.  Tea,  5s.  per  pound;  Sugar,  4  pence. 

Prob.  4.  What  fraction  is  that  to  the  numerator  of  which  if 
4  be  added  the  value  is  one  half;  but  if  7  be  added  to  the  de 
nominator,  its  value  is  one  fifth  ?  Ans.  -^. 

Prob.  5.  A  certain  sum  of  money,  put  out  at  simple  interest, 
amounts  in  8  months  to  $1488,  and  in  15  months  it  amounts 
to  $1530.  What  is  the  sum  and  rate  per  cent.  ? 

Prob.  6.  A  sum  of  money  put  out  at  simple  interest  amounts 
in  m  months  to  a  dollars,  and  in  n  months  to  b  dollars.  Ke- 
quired  the  sum  and  rate  per  cent. 

na—mb     .,  b  —  a 


Ans.  The  sum  is 


n  —  m 


the  rate  is  1200  x 


Prob.  7.  There  is  a  number  consisting  of  two  digits,  the  sec 
ond  of  which  is  greater  than  the  first  ;  and  if  the  number  be 
divided  by  the  sum  of  its  digits,  the  quotient  is  4;  but  if  the 
digits  be  inverted,  and  that  number  be  divided  by  a  number 


110  ALGEBRA. 

greater  by  two  than  the  difference  of  the  digits,  the  quotient  is 
14.  Kequired  the  number. 

Let  x  represent  the  left-hand  digit,  and  y  the  right-hand 
digit. 

Then,  since  x  stands  in  the  place  of  tens,  the  number  will 
be  represented  by  IQx+y. 

Hence,  by  the  first  condition, 

=4; 


x+y 
by  the  second  condition, 


Whence  x—  4,  y=8,  and  the  required  number  is  48. 

Prob.  8.  A  boy  expends  thirty  pence  in  apples  and  pears, 
buying  his  apples  at  4  and  his  pears  at  5  for  a  penny,  and  aft 
erward  accommodates  his  friend  with  half  his  apples  and  one 
third  of  his  pears  for  13  pence.  How  many  did  he  buy  of 
each? 

Prob.  9.  A  father  leaves  a  sum  of  money  to  be  divided 
among  his  children  as  follows:  the  first  is  to  receive  $300  and 
the  sixth  part  of  the  remainder  ;  the  second,  $600  and  the 
sixth  part  of  the  remainder;  and,  generally,  each  succeeding 
one  receives  $300  more  than  the  one  immediately  preceding, 
together  with  the  sixth  part  of  what  remains.  At  last  it  is 
found  that  all  the  children  receive  the  same  sum.  What  was 
the  fortune  left,  and  the  number  of  children  ? 

Ans.  The  fortune  was  $7500,  and  the  number  of  children  5. 

Prob.  10.  A  sum  of  money  is  to  be  divided  among  several 
persons  as  follows:  the  first  receives  a  dollars,  together  with 
the  nth  part  of  the  remainder;  the  second,  2a,  together  with  the 
nth  part  of  the  remainder;  and  each  succeeding  one  a  dollars 
more  than  the  preceding,  together  with  the  nth  part  of  the  re 
mainder  ;  and  it  is  found  at  last  that  all  have  received  the  same 
Bum.  What  was  the  amount  divided,  and  the  number  of  per 
sons?  Ans.  The  amount  was  a(n  —  I)3; 

the  number  of  persons  =r?  —  1. 

Prob.  11.  A  wine-dealer  has  two  kinds  of  wine.    If  he  mixes 


EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY.  Ill 

9  quarts  of  the  poorer  with  7  quarts  of  the  better,  he  can  sell 
the  mixture  at  55  cents  per  quart;  but  if  he  mixes  3  quarts  of 
the  poorer  with  5  quarts  of  the  better,  he  can  sell  the  mixture 
at  58  cents  per  quart.  What  was  the  cost  of  a  quart  of  each 
kind  of  wine? 

Ans.  48  cents  for  the  poorer,  and  64  for  the  better. 

Prob.  12.  A  person  owes  a  certain  sum  to  two  creditors.  At 
one  time  he  pays  them  $530,  giving  to  one  four  elevenths  of 
the  sum  which  is  due,  and  to  the  other  $30  more  than  one 
sixth  of  his  debt  to  him.  At  a  second  time  he  pays  them  $420, 
giving  to  the  first  three  sevenths  of  what  remains  due  to  him, 
and  to  the  other  one  third  of  what  remains  due  to  him.  What 
were  the  debts  ?^_ 

Prob.  13.  If  A  and  B  together  can  perform  a  piece  of  work 
in  12  days,  A  and  C  together  in  15  days,  and  B  and  C  in  20 
days,  how  many  days  will  it  take  each  person  to  perform  the 
same  work  alone? 

This  problem  is  readily  solved  by  first  finding  in  what  time 
they  could  finish  it  if  all  worked  together. 

Prob.  14.  If  A  and  B  together  can  perform  a  piece  of  work 
in  a  days,  A  and  C  together  in  b  days,  and  B  and  C  in  c  days, 
how  many  days  will  it  take  each  person  to  perform  the  same 

work  alone?  rt  ,  0  7 

.    .  2abc  -p,  .  2abc         , 

Ans.  A  in  —  —,  --  7  days-;  B  in  -  days; 

ac  +  bc—ab  ab  +  bc—ac 


n  •  -, 

C  m  -j—      —  j-  days. 

ab  -\-ac-  be 

Prob.  15.  A  merchant  has  two  casks,  each  containing  a  cer 
tain  quantity  of  wine.  In  order  to  have  an  equal  quantity  in 
each,  he  pours  out  of  the  first  cask  into  the  second  as  much  as 
the  second  contained  at  first;  then  he  pours  from  the  second 
into  the  first  as  much  as  was  left  in  the  first  ;  and  then  again 
from  the  first  into  the  second  as  much  as  was  left  in  the  second, 
when  there  are  found  to  be  a  gallons  in  each  cask.  How  man;y 
gallons  did  each  cask  contain  at  first  ? 

lla       ,  Sa 
Ans.  -rr-  and  -—. 
'4l  & 


112  ALGEBRA. 

Prob.  16.  A  laborer  is  engaged  for  n  days  on  condition  that 
he  receives  p  pence  for  every  day  he  works,  and  pays  q  pence 
for  every  day  he  is  idle.  At  the  end  of  the  time  he  receives  a 
pence.  How  many  days  did  he  work,  and  how  many  was  he 
idle? 

Ans.  He  worked  — days,  and  was  idle  -£• days. 

p  +  q  p  +  q 

Prob.  17.  A  certain  number  consisting  of  two  digits  contains 
the  sum  of  its  digits  four  times,  and  their  product  three  times. 
What  is  the  number? 

Prob.  18.  A  father  says  to  his  two  sons,  of  whom  one  was 
four  years  older  than  the  other,  In  two  years  my  age  will  be 
double  the  sum  of  your  ages ;  but  6  years  ago  my  age  was  6 
times  the  sum  of  your  ages.  How  old  was  the  father  and  each 
of  the  sons? 
Ans.  The  father  was  42,  one  son  11,  and  the  other  7  years  old. 

Prob.  19.  It  is  required  to  divide  the  number  96  into  three 
parts  such  that  if  we  divide  the  first  by  the  second  the  quo 
tient  shall  be  2,  with  3  for  a  remainder ;  but  if  we  divide  the 
second  by  the  third,  the  quotient  shall  be  4,  with  5  for  a  re 
mainder.  What  are  the  three  parts?  Ans.  61,  29,  and  6. 

Prob.  20.  Each  of  seven  baskets  contains  a  certain  number 
of  apples.  I  transfer  from  the  first  basket  to  each  of  the  other 
six  as  many  apples  as  it  previously  contained ;  I  next  trans 
fer  from  the  second  basket  to  each  of  the  other  six  as  many 
apples  as  it  previously  contained,  and  so  on  to  the  last  basket, 
when  it  appeared  that  each  basket  contained  the  same  number 
of  apples,  viz.,  128.  How  many  apples  did  each  basket  contain 
before  the  distribution? 

Ans.  The  first  449,  the  second  225,  the  third  113,  the  fourth 
57  the  fifth  29,  the  sixth  15,  and  the  seventh  8  apples. 

155.  When  we  have  only  one  equation  containing  more  than 
one  unknown  quantity,  we  can  generally  solve  the  equation  in 
an  infinite  number  of  ways.  For  example,  if  a  problem  involv 
ing  two  unknown  quantities  (x  and  y)  leads  to  the  singh  equa 
tion 


-EQUATIONS  WITH  MORE  THAN  ONE  UNKNOWN  QUANTITY. 

we  may  ascribe  any  value  we  please  to  x,  and  then  determine 
the  corresponding  value  of  y.  Such  a  problem  is  called  inde 
terminate.  An  indeterminate  problem  is  one  which  admits  of  an 
indefinite  number  of  solutions. 

156.  If  we  had  two  equations  containing  three  unknown  quan 
tities,  we  could,  in  the  first  place,  eliminate  one  of  the  unknown 
quantities  by  means  of  the  proposed  equations,  and  thus  obtain 
one  equation  containing  two  unknown  quantities,  which  would 
be  satisfied  by  an  infinite  number  of  systems  of  values.    There 
fore,  in  order  that  a  problem  may  be  determinate,  its  enuncia 
tion  must  contain  as  many  different  conditions  as  there  are  un 
known  quantities,  and  each  of  these  conditions  must  be  express 
ed  by  an  independent  equation. 

157.  Equations  are  said  to  be  independent  when  they  express 
conditions  essentially  different,  and  dependent  when  they  express 
the  same  conditions  under  different  for  7ns. 

Thus  \A   ~*y~       \  are  independent  equations. 
I  2x+y=W  ) 

But  •!     '     9  _  14.C  are   n°t  independent,  because  the  one 
may  be  deduced  from  the  other. 

158.  If,  on  the  contrary,  the  number  of  independent  equa 
tions  exceeds  the  number  of  unknown  quantities,  these  equa 
tions  will  be  contradictory. 

For  example,  let  it  be  required  to  find  two  numbers  such  that 
their  sum  shall  be  8,  their  difference  2,  and  their  product  20. 
From  these  conditions  we  derive  the  following  equations  : 


ay  =20. 

From  the  first  two  equations  we  find 
tc=5  and  ?/=3. 

Hence  the  third  condition,  which  requires  that  their  product 
ehall  be  equal  to  20,  can  riot  be  fulfilled. 


114  ALGEBBA. 


CHAPTEK  X. 

DISCUSSION  OF  PROBLEMS  INVOLVING   SIMPLE   EQUATIONS. — • 
INEQUALITIES. 

159.  To  discuss  a  problem  or  an  equation  is  to  determine  the 
values  which  the  unknown  quantities  assume  for  particular  hy 
potheses  made  upon  the  values  of  the  given  quantities,  and  to 
interpret  the  peculiar  results  obtained.     We  have  seen  that  if 
the  sum  of  two  numbers  is  represented  by  a,  and  their  differ 
ence  by  5,  the  greater  number  will  be  expressed  by  — — ,  and 
the  less  by  — — .    Here  a  and  b  may  have  any  values  whatever, 

JU 

and  still  these  formulas  will  always  hold  true.  It  frequently 
happens  that,  by  attributing  different  values  to  the  letters  which 
represent  known  quantities,  the  values  of  the  unknown  quanti 
ties  assume  peculiar  forms,  which  deserve  consideration. 

160.  We  may  obtain  jive  species  of  values  for  the  unknown 
quantity  in  a  problem  of  the  first  degree: 

1st.  Positive  values. 

2d.  Negative  values.  ~ 

3d.  Values  of  the  form  of  zero,  or  -r. 

A 

4th.  Values  of  the  form  of  — . 

5th.  Values  of  the  form  of  -z. 

We  will  consider  these  five  cases  in  succession. 

161.  1st.  Positive  values  are  generally  answers  to  problems  in 
the  sense  in  which  they  are  proposed.     Nevertheless,  all  posi 
tive  values  will  not  always  satisfy  the  enunciation  of  a  prob 
lem.    For  example,  a  problem  may  require  an  answer  in  whole 


DISCUSSION  OF   PROBLEMS.  115 

numbers,  in  which  case  a  fractional  value  of  the  unknown  quan 
tity  is  inadmissible.  Thus,  in  Prob.  17,  page  93,  it  is  implied 
that  the  value  of  x  must  be  a  whole  number,  although  this  con 
dition  is  not  expressed  in  the  equations.  We  might  change 
the  data  of  the  problem,  so  as  to  obtain  a  fractional  value  of 
x,  which  would  indicate  an  impossibility  in  the  problem  pro 
posed.  Problem  43,  page  97,  is  of  the  same  kind  ;  also  Prob, 
7,  page  109. 

If  the  value  obtained  for  the  unknown  quantity,  even  when 
positive,  does  not  satisfy  all  the  conditions  of  the  problem,  the 
problem  is  impossible  in  the  form  proposed. 

162.  2d.  Negative  values. 

Let  it  be  proposed  to  find  a  number  which,  added  to  the 
number  6,  gives  for  a  sum  the  number  a.  Let  x  denote  the  re 
quired  number;  then,  by  the  conditions  of  the  problem, 


whence  cc  =  a  —  b. 

This  formula  will  give  the  value  of  x  corresponding  to  any 
assigned  values  of  a  and  b. 

For  example,  if          a  =  7  and  b  =  4:, 
then  x  =  7—  4  =  3, 

a  result  which  satisfies  the  conditions. 

But  suppose  that        a  —5  and  b  =  8, 
then  x  —  5—  8=—  3. 

We  thus  obtain  for  x  a  negative  value.  How  is  it  to  be  in 
terpreted  ? 

By  referring  to  the  problem,  we  see  that  it  now  reads  thus: 
What  number  must  be  added  to  8  in  order  that  the  sum  rn;iy 
be  5?  It  is  obvious  that  if  the  word  added  and  the  word  sum 
are  to  retain  their  arithmetical  meanings,  the  proposed  problem 
is  impossible.  Nevertheless,  if  in  the  equation  8  +  x  =  5  we 
substitute  for  +x  its  value  —3,  it  becomes 

8-3  =  5, 

an  identical  equation;  that  is,  8  diminished  by  3  is  equal  to  5, 
or  5  may  be  regarded  as  the  algebraic  sum  of  8  and  —3. 

The  negative  result,  x=—  3,  indicates  that  the  problem,  in  a 


116  ALGEBRA. 

strictly  arithmetical  sense,  is  impossible ;  but,  taking  this  value 
of  x  with  a  contrary  sign,  we  see  that  it  satisfies  the  enunciation 
when  modified  as  follows :  What  number  must  be  subtracted 
from  8  in  order  that  the  difference  may  be  5  ?  The  second 
enunciation  differs  from  the  first  only  in  this,  that  we  put  sub 
tract  for  add,  and  difference  for  sum. 
If  we  wish  to  solve  this  new  equation  directly,  we  shall  have 

8-z=5; 
whence  x=S— 5,  or  3. 

163.  For  another  example,  take  Problem  50,  page  98.  The 
age  of  the  father  being  represented  by  a,  and  that  of  the  son 

by  Z>,  then  -  — -  will  represent  the  number  of  years  before  the 

age  of  the  father  will  be  n  times  that  of  the  son. 
Thus,  suppose      a =54,  5—9,  and  ?z=4; 

54-36     18 
then  x  =  —^-  =  -g  =  6. 

This  value  of  x  satisfies  the  conditions  understood  arithmet 
ically ;  for  if  the  father  was  54  years  old,  and  the  son  9  years, 
then  in  6  years  more  the  age  of  the  father  will  be  60  and  the 
son  15 ;  and  we  see  that  60  is  4  times  15. 

But  suppose       a =45,  £=15,  and  n=4; 

45-60     -15 
then  x— — - — = — —-  =  —5. 

O  O 

Here  again  we  obtain  a  negative  result.  How  are  we  to  in 
terpret  it? 

By  referring  to  the  problem,  we  see  that  the  age  of  the  son 
is  already  more  than  one  fourth  that  of  the  father,  so  that  the 
time  required  is  already  past  by  five  years.  The  problem,  if 
taken  in  a  strictly  arithmetical  meaning,  is  impossible.  But 
let  us  modify  the  enunciation  as  follows : 

The  age  of  the  father  is  45  years;  the  son's  age  is  15  years; 
how  many  years  since  the  age  of  the  father  was  four  times  that 
of  his  son  ? 


DISCUSSION   OF  PROBLEMS.  117 

The  equation  corresponding  to  this  new  enunciation  is 


4 

whence  60—  4x=45—  x;   and  x—5, 

a  result  which  satisfies  the  modified  problem  taken  in  its  arith 
metical  sense. 

From  this  discussion  we  derive  the  following  general  prin 
ciples  : 

1st.  A  negative  result  found  for  the  unknown  quantity  in  a  prob 
km  of  the  first  degree  indicates  that  the  probkm  is  impossible,  if 
understood  in  its  strict  arithmetical  sense. 

2d.  This  negative  value,  taken  with  a  contrary  sign,  may  be  re 
garded  as  the  answer  to  a  probkm  whose  enunciation  only  differs 
from  that  of  the  proposed  probkm  in  this,  that  certain  quantities 
which  were  ADDED  should  have  been  SUBTRACTED,  and  vice  versa. 

164.  In  what  case  would  the  value  of  the  unknown  quantity 
in  Prob.  20,  page  94,  be  negative?  Ans.  When  n>m. 

Thus,  let       m  —  20,  n  =  25,  and  a  =  60  miles; 

then  x-     6°      -6°-      12 

~20-25~-5~ 

To  interpret  this  result,  observe  that  it  is  impossible  that  the 
second  train,  which  moves  the  slowest,  should  overtake  the  first. 
At  the  time  of  starting,  the  distance  between  them  was  60  miles, 
and  each  subsequent  hour  the  distance  increases.  If,  however, 
we  suppose  the  two  trains  to  have  been  moving  uniformly  along 
an  endless  road,  it  is  obvious  that  at  some  former  time  they  must 
have  been  together. 

This  negative  result  indicates  that  the  problem  is  impossible 
if  understood  in  its  strict  arithmetical  sense.  But  if  the  prob 
lem  had  been  stated  thus  : 

Two  trains  of  cars,  60  miles  apart,  are  moving  in  the  same 
direction,  the  forward  one  25  miles  per  hour,  the  other  20. 
How  long  since  they  were  together? 

The  problem  would  have  furnished  the  equation 


whenoe  x=  +12. 


118  ALGEBKA. 

If  we  wish  to  include  both  of  these  cases  in  the  same  enun 
ciation,  the  question  should  be,  Required  the  time  of  their  being 
together,  leaving  it  uncertain  whether  the  time  w as  past  or  future* 

EXAMPLES. 

1.  What  number  is  that  whose  fourth  part  exceeds  its  third; 
part  by  16?  Ans.  —192. 

How  should  the  enunciation  be  modified  in  order  that  the 
result  may  be  positive? 

2.  The  sum  of  two  numbers  is  2,  and  their  difference  8- 
What  are  those  numbers?  Ans.  —  3  and  +5. 

How  should  the  enunciation  be  modified  in  order  that  both 
results  may  be  positive  ? 

3.  What  fraction  is  that  from  the  numerator  of  which  if  4 
be  subtracted  the  value  is  one  half,  but  if  7  be  subtracted 

from  the  denominator  its  value  is  one  fifth  ?  —5 

Ans. 


-18 

How  should  the  enunciation  be  modified  in  order  that  the 
problem  may  be  possible  in  its  arithmetical  sense? 

4.  Find  two  numbers  whose  difference  is  6,  such  that  four 
times  the  less  may  exceed  five  times  the  greater  by  12. 

Ans.  —42  and  —36. 

Change  the  enunciation  of  the  problem  so  that  these  num 
bers,  taken  with  the  contrary  sign,  may  be  the  answers  to  the 
modified  problem, 

~^ 

165.  3d.  We  may  obtain  for  the  unknown  quantity  values  of 

the  form  of  zero,  or  -r. 
j\. 

In  what  case  would  the  value  of  the  unknown  quantity  in 
Prob.  20,  page  94,  become  zero,  and  what  would  this  value 
signify  ? 

Ans.  This  value  becomes  zero  when  a  =  0,  which  signifies 
that  the  two  trains  are  together  at  the  outset. 

In  what  case  would  the  value  of  the  unknown  quantity  in 
Prob.  50,  page  98,  become  zero,  and  what  would  this  value 
signify? 


DISCUSSION   OF   PROBLEMS.  119 

Ans.  When  a  — rib,  which  signifies  that  the  age  of  the  father 
is  now  n  times  that  of  the  son. 

In  what  case  would  the  values  of  the  unknown  quantities  in 
Prob.  38,  page  96,  become  zero,  and  what  would  these  values 
signify  ? 

When  a  problem  gives  zero  for  the  value  of  the  unknown 
quantity,  this  value  is  sometimes  applicable  to  the  problem, 
and  sometimes  it  indicates  an  impossibility  in  the  proposed 
question. 

166.  4th.  We  may  obtain  for  the  unknown  quantity  values 
j^  

of  the  form  of  — . 

In  what  case  does  the  value  of  the  unknown  quantity  in 

A 

Prob.  20,  page  94,  reduce  to  -^-,  and  how  shall  we  interpret 
this  result?  Ans.  When  m  =  n. 

On  referring  to  the  enunciation  of  the  problem,  we  see  that 
it  is  absolutely  impossible  to  satisfy  it ;  that  is,  there  can  be  no 
point  of  meeting;  for  the  two  trains, being  separated  by  the  dis 
tance  a,  and  moving  equally  fast,  will  always  continue  at  the 

same  distance  from  each  other.    The  result  ^  may  then  be  re 
garded  as  indicating  an  impossibility. 

The  symbol  Q  is  sometimes  employed  to  represent  infinity, 
and  for  the  following  reason : 

•If  the  denominator  of  a  fraction  is  made  to  diminish,  while 
the  numerator  remains  unchanged,  the  value  of  the  fraction 
must  increase. 

For  example,  let          m— n  =  0.01 ; 

then  a=— ^-=-^ 

m  —  n     .01 

Let  m— n  —  0.0001; 

then  x=,_JL_=_JjL.  =  i0 

m^-n     .0001 

Hence,  if  the  difference  in  the  rates  of  motion  is  not  zero,  the 


120  ALGEBRA. 

two  trains  must  meet,  and  the  time  will  become  greater  and 
greater  as  this  difference  is  diminished.  If,  then,  we  suppose 
this  difference  to  be  less  than  any  assignable  quantity,  the  time 

represented  by  -      -  will  be  greater  than  any  assignable  quantity. 

rffL  —  Tif 

Hence  we  infer  that  every  expression  of  the  form  —  found 

for  the  unknown  quantity  indicates  the  impossibility  of  satis- 
fying  the  problem,  at  least  in  finite  numbers. 

In  what  case  would  the  value  of  the  unknown  quantity  in 

Prob.  10,  page  92,  reduce  to  the  form  -^,  and  how  shall  we  in 
terpret  this  result? 

167.  The  s^ymbol  0,  called  zero,  is  sometimes  used  to  denote 
the  absence  of  value,  and  sometimes  to  denote  a  quantity  less 
than  any  assignable  value. 

The  symbol  oo ,  called  infinity,  is  used  to  denote  a  quantity 
greater  than  any  assignable  value.  A  line  produced  beyond  any 
assignable  limit  is  said  to  be  of  infinite  length ;  and  time  ex 
tended  beyond  any  assignable  limit  is  called  infinite  duration. 

We  have  seen  that  when  the  denominator  of  the  fraction 

^2~  becomes  less  than  any  assignable  quantity,  the  value  of 
the  fraction  becomes  greater  than  akf  assignable  quantity. 
Hence  we  conclude  that  ^  —  oo; 

that  is,  a  finite  quantity  divided  by  tero  is  an  expression  for  in 
finity. 

Also,  if  the  denominator  of  a  fraction  be  made  to  increase 
while  the  numerator  remains  unchanged,  the  value  of  the  frac 
tion  must  diminish;  and  when  the  denominator  becomes  greater 
than  any  assignable  quantity,  the  value  of  the  fraction  must  be 
come  less  than  any  assignable  quantity.  Hence  we  conclude  that 

-=0; 

00 

that  is,  a  finite  quantity  divided  ly  infinity  is  an  expression  for 
zero 


DISCUSSION   OF   PEOBLEMS. 


121 


168.  5th.  We  may  obtain  for  the  unknown  quantity  values 
of  the  form  of  ^. 

In  what  case  does  the  value  of  the  unknown  quantity  in 
Prob.  20,  page  94,  reduce  to  77,  and  how  shall  we  interpret  this 
lesult?  Ans.  When  a  =  0,  and  m  —  n. 

To  interpret  this  result,  let  us  recur  to  the  enunciation,  and 
observe  that,  since  a  is  zero,  both  trains  start  from  the  same 
point;  and  since  they  both  travel  at  the  same  rate,  they  will 
always  remain  together ;  and,  therefore,  the  required  point  of 
meeting  will  be  any  where  in  the  road  traveled  over.  The 
problem,  then,  is  entirely  indeterminate,  or  admits  of  an  infinite 

number  of  solutions ;  and  the  expression  ~  may  represent  any 
finite  quantity. 
We  infer,  therefore,  that  an  expression  of  the  form  of  Q  found 

for  the  unknown  quantity  generally  indicates  that  it  may  have 
any  value  whatever.  In  some  cases,  however,  this  value  is 
subject  to  limitations. 

In  what  case  would  the  values  of  the  unknown  quantities 

in  Prob.  44,  page  97,  reduce  to  ^,  and  how  would  they  satisfy 
the  conditions  of  the  problem?  Ans.  When  a  =  6  =  c, 

which  indicates  that  the  coins  are  all  of  the  same  value.     B 
might  therefore  be  paid  in  either  kind  of  coin ;  but  there  is  a 
limitation,  viz.,  that  the  value  of  the  coins  must  be  one  dollar. 
In  what  case  do  the  values  of  the  unknown  quantities  in 

Prob.  38,  page  96,  reduce  to  -r,  and  how  shall  we  interpret 
his  result? 

169.  The  expression  -  may  be  conceived  to  result  from  a 

fraction  whose  numerator  and  denominator  both  diminish  si 
multaneously,  but  in  such  a  manner  as  to  preserve  the  same 
relative  value.  If  both  numerator  and  denominator  of  a  frac 
tion  are  divided  by  the  same  quantity,  its  value  remains  ur> 

K 


122  ALGEBRA. 

changed.    Hence,  if  ~  represent  any  fraction,  we  may  conceive 

both  numerator  and  denominator  to  be  divided  by  10,  100, 
1000,  etc.,  until  each  becomes  less  than  any  assignable  quanti 
ty,  or  0.  The  fraction  then  reduces  to  the  form  of  ^  but  the 
ralue  of  the  fraction  has  throughout  remained  unchanged. 

For  example,  we  may  suppose  the  numerator  to  represent 
the  circumference  of  a  circle,  and  the  denominator  to  represent 
its  diameter.  The  value  of  the  fraction  in  this  case  is  known 
to  be  3.1416.  If  now  we  suppose  the  circle  to  diminish  until 
it  becomes  a  mere  point,  the  circumference  and  diameter  both 
become  zero,  but  the  value  of  the  fraction  has  throughout  re 
mained  the  same.  Hence,  in  this  case,  we  have 


Again,  suppose  the  numerator  to  represent  the  area  of  a  cir 
cle,  and  the  denominator  the  area  of  the  circumscribed  square; 
then  the  value  of  the  fraction  becomes  .7854C  But  this  value 
remains  unchanged,  although  the  circle  may  be  supposed  to 
diminish  until  it  becomes  a  mere  point.  Hence,  in  this  case, 
we  have  0 


Hence  we  conclude  that  the  symbol  -^  may  represent  any 
finite  quantity. 

So,  also,  we  may  conceive  both  numerator  and  denominator 
of  a  fraction  to  be  multiplied  by  10,  100,  1000,  etc.,  until  each* 
becomes  greater  than  any  assignable  quantity  ;  the  fraction 

then  reduces  to  the  form  of  —  .     Hence  we  conclude  that  the 

oo  °° 

symbol  —  may  also  represent  any  finite  quantity. 

INEQUALITIES. 

170.  An  inequality  is  an  expression  denoting  that  one  quan 
tity  »is  greater  or  less  than  another.  Thus  Sx  >  2a5  denotes 
that  three  times  the  quantity  x  is  greater  than  twice  the  prod- 
act  of  the  quantities  a  and  b. 


INEQUALITIES.  123 

171.  In  treating  of  inequalities,  the  terms  greater  and  less 
must  be  understood  in  their  algebraic  sense  ;  that  is,  a  negative 
quantity  standing  alone  is  regarded  as  less  than  zero;  and  of 
two  negative  quantities,  that  which  is  numerically  the  greatest 
is  considered  as  the  least;  for  if  from  the  same  number  we  sub 
tract  successively  numbers  larger  and  larger,  the  remainders 
must  continually  diminish.    Take  any  number,  5  for  example, 
and  from  it  subtract  successively  1,  2,  3,  4,  5,  6,  7,  8,  9,  etc., 
we  obtain 

5-1,  5-2,  5-3,  5-4,  5-5,  5-6,  5-7,  5-8,  5-9,  etc., 
or,  reducing,  we  have 

4,  3,  2,  1,  0,  -1,  -2,  -3,  -4,  etc. 

Hence  we  see  that  —  1  should  be  regarded  as  less  than  zero  ; 
—2  less  than  —1;   —3  less  than  —2,  etc. 

172.  Tw®  inequalities  are  said  to  subsist  in  the  same  sense 
when  the  greater  quantity  stands  at  the  left  in  both,  or  at  the 
right  in  both  ;  and  in  a  contrary  sense  when  the  greater  quanti 
ty  stands  at  the  right  in  one  and  at  the  left  in  the  other.    Thus 
9>7  and  7>6,  or  5<8  and  3<4,  are  inequalities  which  sub 
sist  in  the  same  sense;    but  the  inequalities  10>6  and  3<7 
subsist  in  a  contrary  sense. 

173.  Properties  of  Inequalities.  —  1st.  If  the  same  quantity  be 
added  to  or  subtracted  from  each  member  of  an  inequality,  the  re 
suiting  inequality  will  ahuays  subsist  in  the  same  sense. 

Thus,  8>3. 

Adding  5  to  each  member,  we  have 


and  subtracting  5  from  each  member,  we  have 

8_5>3-5. 
Again,  take  the  inequality 

-3<-2. 
Adding  6  to  each  member,  we  have 

_3  +  6<-2  +  6,  or  3<4; 
and  subtracting  6  from  each  member^ 

_3-6<-2-6,  or  -9<-& 


124  ALGEBRA. 

174.  Hence  we  conclude  that  we  may  transpose  a  term  from 
one  member  of  an  inequality  to  the  other,  provided  we  change 
its  sign. 

Thus,  suppose  a2  +  62>3£2-2a2. 

Adding  2a2  to  each  member  of  the  inequality,  it  becomes 


Subtracting  b2  from  each  member,  we  have 

a24-2a2>362-&2, 
or  3a2>262. 

175.  2d.  If  we  add  together  the  corresponding  members  of  two  or 
more  inequalities  which  subsist  in  the  same  sense,  the  resulting  in 
equality  will  always  subsist  in  the  same  sense. 

Thus,  S>4 

4>2 

7>3 
Adding,  we  obtain  16  >  9. 

176.  3d.  If  one  inequality  be  subtracted  from  another  which  sub 
sists  in  the  same  sense,  the  result  will  not  always  be  an  inequality 
subsisting  in  the  same  sense. 

Take  the  two  inequalities   4<7 
and  2<3 

Subtracting,  we  have    4—  2<7—  3,  or  2<4, 

where  the  result  is  an  inequality  subsisting  in  the  same  sense. 

But  take  9<10 

and  6<  8 

Subtracting,  we  have  9  —  6>10—  8,  or  3>2, 

where  the  result  is  an  inequality  subsisting  in  the  contrary  sense. 
We  should  therefore  avoid  as  much  as  possible  the  use  of 
this  transformation,  or,  when  we  employ  it,  determine  in  what 
sense  the  resulting  inequality  subsists. 

177.  4th.  If  we  multiply  or  divide  each  member  of  an  inequality 
"by  the  same  positive  quantity,  the  resulting  inequality  will  subsist 
in  the  same  sense. 


INEQUALITIES.  126 

Thus,  if  a<b, 

then  ma<mb, 

,  a     b 

and  — <— . 

m    m 

Also,  if  —  a>—  b, 

then  —ma>—mb. 

,  a          b 

and  > . 

m        m 

Hence  an  inequality  may  be  cleared  of  fractions.    Thus,  sup 
pose  we  have  a2  —  b2    c2—d2 
~2d~>~3a~' 

Multiplying  each  member  by  6ad,  it  becomes 
3a(a2-b2)>2d(c2-d2). 

178.  5i\i.\Tf  we  multiply  or  divide  each  member  of  an  inequality 
by  the  same  negative  number,  the  resulting  inequality  will  subsist 
in  the  contrary  sense. 

Take,  for  example,  8>7. 

Multiplying  each  member  by  —3,  we  have  the  opposite  in 
equality  —  24  <  —  21. 

So,  also,  15>12. 

Dividing  each  member  by  —  3,  we  have 
-5<-4, 

Therefore,  if  we  multiply  or  divide  the  two  members  of  an 
inequality  by  an  algebraic  quantity,  it  is  necessary  to  ascertain 
whether  the  multiplier  or  divisor  is  negative,  for  in  this  case 
the  resulting  inequality  subsists  in  a  contrary  sense. 

179.  6th.  If  the  signs  of  all  the  terms  of  an  inequality  be  changed, 
the  sign  of  inequality  must  be  reversed. 

For  to  change  all  the  signs  is  equivalent  to  multiplying  each 
member  of  the  inequality  by  —1. 

180.  Reduction  of  Inequalities. — The  principles  now  establish 
ed  enable  us  to  reduce  an  inequality  so  that  the  unknown  quan 
tity  may  stand  alone  as  one  member  of  the  inequality.      The 


126  ALGEBRA. 

other   member  will  then  denote  one  limit  of  the  unknown 
quantity. 

EXAMPLES. 

1.  Find  a  limit  of  x  in  the  inequality 

7      5rc    95     0 

6~T<12~ 
Multiplying  each  member  by  12,  we  have 

14-15x<95-24x. 
Transposing,  9x<81. 

Dividing,  •  .      x<9. 


2. 

3.   S,-2>- 


_    x     x— 

5.   -  ---  ^-<x  --  -  —  .  Ans. 

D  1O 


6.  Given 


to  find  the  limits  of  x. 


Aiis. 


7.  A  man,  being  asked  how  many  dollars  he  gave  for  his 
watch,  replied,  If  you  multiply  the  price  by  4,  and  to  the  prod 
uct  add  60,  the  sum  will  exceed  256;  but  if  you  multiply  the 
price  by  3,  and  from  the  product  subtract  40,  the  remainder 
will  be  less  than  113.     Eequired  the  price  of  the  watch. 

8.  What  number  is  that  whose  half  and  third  part  added 
together  are  less  than  105 ;  but  its  half  diminished  by  its  fifth 
part  is  greater  than  33? 

9.  The  double  of  a  number  diminished  by  6  is  greater  than 
22,  and  triple  the  number  diminished  by  6  is  less  than  double 
the  number  increased  by  10.     Eequired  the  number. 


INVOLUTION.  127 


CHAPTER  XL 

INVOLUTION. 

181.  A  power  of  a  quantity  is  the  product  obtained  by  tak 
ing  that  quantity  any  number  of  times  as  a  factor. 

Thus  the  first  power  of  3  is  3  ; 

the  second  power  of  3  is  3  x  3,  or  9  ; 

the  fourth  power  of  3  is  3  x  3  x  3  x  3,  or  81,  etc. 
Involution  is  the  process  of  raising  a  quantity  to  any  power. 

182.  A  power  is  indicated  by  means  of  an  exponent.     The 
exponent  i^a  number  or  letter  written  a  little  above  a  quantity 
to  the  right,  and  shows  how  many  times  that  quantity  is  taken 
as  a  factor. 

Thus  the  first  power  of  a  is  a1,  where  the  exponent  is  1, 
which,  however,  is  commonly  omitted. 

The  second  power  of  a  is  a  x  a,  or  a2,  where  the  exponent  2 
denotes  that  a  is  taken  twice  as  a  factor  to  produce  the  power 
aa. 

The  third  power  of  a  is  a  x  a  x  a,  or  a3,  where  the  exponent 
3  denotes  that  a  is  taken  three  times  as  a  factor  to  produce  the 
power  aaa. 

The  fourth  power  of  a  is  a  x  a  x  a  X  a,  or  a4. 

Also  the  rcth  power  of  a  is  a  x  a  x  a  x  a,  etc.,  or  a  repeated 
as  a  factor  n  times,  and  is  written  an. 

The  second  power  is  commonly  called  the  square,  and  the 
third  power  the  cube. 

183.  Exponents  may  be  applied  to  polynomials  as  well  as 
to  monomials. 

Thus  (a+£  +  c)3  is  the  same  as 


or  the  third  power  of  the  entire  expression  a~\-l> 


128  ALGEBRA. 

Powers  of  Monomials. 

184.  Let  it  be  required  to  find  the  third  power  or  cube  of 
2a3b2. 

According  to  the  rule  for  multiplication,  we  have 

(2aW)3  =  2M2  x  2a3b2  x  2a3b2=2  x  2  x  2a3a3aW62=8a966. 

In  a  similar  manner  any  monomial  may  be  raised  to  any 
power. 

Hence,  to  raise  a  monomial  to  any  power,  we  have  the  fol 
lowing 

RULE. 

Raise  the  numerical  coefficient  to  the  required  power,  and  multi 
ply  the  exponent  of  each  of  the  letters  by  the  exponent  of  the  re 
quired  power. 

185.  Sign  of  the  Power. — With  respect  to  the  signs,  it  is  ob 
vious  from  the  rules  for  multiplication  that  if  the  given  mono 
mial  be  positive,  all  of  its  powers  are  positive ;  but  if  the  mo 
nomial  be  negative,  its  square  is  positive,  its  cube  negative,  its 
fourth  power  positive,  and  so  on. 

Thus  —ax—  a=-fa2, 

—  ax  —ax  —  a——  a3, 
—  ax—  ax—  ax—  a=+a4, 
—  ax  —  ax  —  ax  —  ax  —  a  =  —  a5,  etc. 

In  general,  any  even  power  of  a  negative  quantity  is  positive, 
and  every  odd  power  negative  ;  but  all  powers  of  a  positive  quan 
tity  are  positive. 

EXAMPLES. 

1.  Find  the  square  of  lla3M2.  Ans.  121a66W4. 

2.  Find  the  square  of  —  lSx2yz3. 
8.  Find  the  cube  of  7ab2x2. 

4.  Find  the  cube  of  -8xy2z3. 

5.  Find  the  fourth  power  of  4a52c3. 

6.  Find  the  fourth  power  of  —5a3l2x. 

7.  Find  the  fifth  power  of  2aZ>3x2. 

8.  Find  the  fifth  power  of  -3ab2x*. 

9.  Find  the  sixth  power  of  3ab2x*. 


INVOLUTION. 


129 


10.  Find  the  sixth  power  of  —  2azb3x*. 

11.  Find  the  seventh  power  of  2a2x3y. 

12.  Find  the  rath  power  of  abV. 

186.  Powers  of  Fractions.—  Let  it  be  required  to  find  the 

.,  .   -,  f  2ab2 

third  power  of  -—  . 

From  the  rule  for  the  multiplication  of  fractions,  we  have 


33         3c  ~27c3' 

In  a  similar  manner  any  fraction  may  be  raised  to  any 
power.  Hence,  to  raise  a  fraction  to  any  power,  we  have  the 
following 

EULE. 

Raise  both  numerator  and  denominator  to  the  required  power. 
\ 

EXAMPLES. 

-  3a&2c3 
1.  Find  the  square  of  .  Ans. 


2.  Find  the  square  of  —  -jr 

_  fiox2 

3.  Find  the  cube  of  -—  —  - 

oran 

o 

* 

4.  Find  the  cube  of  — 


A  /y  nr^y/ 

5,  Find  the  fourth  power  of  —  jr~-. 

—  36r7i 

6.  Find  the  fourth  power  of  —  p-r 


7.  Find  the  fifth  power  of  - 


\C1       A    ^ 

8.  Find  the  fifth  power  of 


9.  Find  the  sixth  power  of 

F2 


130  ALGEBRA. 

187.  Negative  Exponents. — The  rule  of  Art.  184,  for  raising 
a  monomial  to  any  power,  holds  true  when  the  exponents  of  any 
of  the  letters  are  negative,  and  also  when  the  exponent  of  the  re 
quired  power  is  negative. 

Let  it  be  required  to  find  the  square  of  a~3.  This  expres 
sion  may  be  written  -3,  which,  raised  to  the  second  power,  be 
comes  -g,  or  a~6,  the  same  result  as  would  be  obtained  by 
multiplying  the  exponent  —3  by  2. 

Also,  let  it  be  required  to  find  that  power  of  2am2  whose 
exponent  is  —3. 

The  expression  (2am2)-3  may  be  written   77- — r-^,  which 
r  (2am2)3 

equals        3    6.     Transferring  the  factors  to  the  numerator,  we 
have  2-3GT37ft-6,  or  -|a~3m~6. 

EXAMPLES. 

Find  the  -value  of  each  of  the  following  expressions. 
1.    (3a26~4)2.  Ans. 

2/fj    273    4    \?  A 

,    ^/a    0' c    x] .  ^Lns. 

3.  (3ab'2x~ly~z)~2.  Ans. 

4.  (-4a2x-y)-2.  Ans. 

5.  ( — 6ab~5x~2)3' 
Q  f 3a2x~3^2)~3. 

8.  (- 

9.  (- 
10.   (- 


188.  Powers  of  Polynomials. — A  polynomial  may  be  raised 
to  any  power  by  the  process  of  continued  multiplication.  If 
the  quantity  be  multiplied  by  itself,  the  product  will  be  the 
second  power;  if  the  second  power  be  multiplied  by  the  orig 
inal  quantity,  the  product  will  be  the  third  power,  and  so  on. 
Hence  we  have  the  following 


INVOLUTION.  131 


KULE. 

Multiply  the  uanlity  l>y  itself  until  it  has  been  taken  as  a  facto 
as  many  times  as  there  are  units  in  the  exponent  of  the  required 
power. 

EXAMPLES. 

1.  Find  the  square  of  2a+362.        Ans.  4a2+12a&2+9&*. 

2.  Find  the  square  of  a+m—n. 

3.  Find  the  cube  of  2a2+3a-l. 

4.  Find  the  cube  of 

5.  Find  the  cube  of 

6.  Find  the  fourth  power  of  a—  b. 

7.  Find  the  fourth  power  of  2a  —  3 

8.  Find  the  fourth  power  of  a3+&3. 

9.  Find  the  fifth  power  of  a—  b. 


1A    .p,.   j    .  ~— 

10.  Find  the  square  of  -  =^.      Ans.  -=- 

2 


a^f-Zabxy  +  bW 

11.  Find  the  cube  of  — '• . 

m— n 

12.  Find  the  cube  of  ^-^-9. 

a— o2 

189.  Square  of  a  Polynomial. — We  have  seen,  Art.  66,  that 
the  square  of  a  binomial  may  be  formed  without  the  labor  of 
actual  multiplication.  The  same  principle  may  be  extended 
to  polynomials  of  any  number  of  terms.  By  actual  multipli 
cation,  we  find  the  square  of  a -\-b-\-c  to  be 


that  is,  the  square  of  a  trinomial  consists  of  the  square  of  each 
term,  together  with  twice  the  product  of  all  the  terms  multiplied  to 
gether  two  and  two. 

In  the  same  manner  we  find  the  square  of  a-\-b-\-c-{-d  to  be 


that  is,  the  square  of  any  polynomial  consists  of  the  square  of  each 
term,  together  with  twice  the  sum  of  the  products  of  all  Hie  terms 
multiplied  together  two  and  two. 


132  ALGEBRA. 

EXAMPLES. 

1.  Find  the  square  of  a-\-b-\-c+d+x. 

2.  Find  the  square  of  a  — b  +  c. 

3.  Find  the  square  of  l  +  2x+3x2. 

4.  Find  the  square  of  1  — x+x2— x3. 

5.  Find  the  square  of  a— 2^+3a6— m. 

6.  Find  the  square  of  1  —  3x+3x2— x3. 

7.  Find  the  square  of  a— 26  +  3c— 4d. 

In  Chapter  XVIII.  will  be  given  a  method  by  which  any 
power  of  a  binomial  may  be  obtained  without  the  labor  of 
multiplication.  "^~ 


EVOLUTION.  133 


CHAPTER  XII. 

EVOLUTION. 

190.  A  root  of  a  quantity  is  one  of  the  equal  factors  which, 
multiplied  together,  will  produce  that  quantity. 

If  a  quantity  be  resolved  into  two  equal  factors,  one  of  them 
is  called  the  square  root. 

If  a 'quantity  be  resolved  into  three  equal  factors,  one  of 
them  is  called  the  cube  root. 

If  a  quantity  be  resolved  into  four  equal  factors,  one  of 
them  is  called  the  fourth  root,  and  so  on. 
t  . 

191.  Evolution  is  the  process  of  extracting  any  root  of  a 
given  quantity. 

Evolution  is  indicated  by  the  radical  sign  y~. 
Thus,  <\fa  denotes  the  square  root  of  a. 

YCL  denotes  the  cube  root  of  a. 

\/a  denotes  the  nib.  root  of  a. 

192.  Surds. — When  a  root  of  an  algebraic  quantity  which 
is  required  can  not  be  exactly  obtained,  it  is  called  an  irration 
al  or  surd  quantity. 

Thus,  Va?  is  called  a  surd,  -v/3  is  also  a  surd,  because  the 
square  root  of  8  can  not  be  expressed  in  numbers  with  perfect 
exactness. 

A  rational  quantity  is  one  which  can  be  expressed  in  finite 
terms,  and  without  any  radical  sign ;  as,  a,  5a2,  etc. 

193.  An  imaginary  root  is  one  which  can  not  be  extracted 
on  account  of  the  sign  of  the  given  quantity.    Thus  the  square 
root  of  — 4  is  impossible,  because  no  quantity  raised  to  an  even 
power  can  produce  a  negative  result. 

A  root  which  is  not  imaginary  is  said  to  be  real. 


134  ALGEBRA. 

Roots  of  Monomials. 

194.  According  to  Art.  184,  in  order  to  raise  a  monomial  tc 
any  power,  we  raise  the  numerical  coefficient  to  the  required 
power,  and  multiply  the  exponent  of  each  of  the  letters  by  the 
exponent  of  the  power  required.     Hence,  conversely,  to  ex 
tract  any  root  of  a  monomial,  we  extract  the  root  of  the  nu 
merical  coefficient,  and  divide  the  exponent  of  each  letter  by 
the  index  of  the  required  root. 

Thus  the  cube  root  of  64a663  is  4a26. 

195.  Sign  of  the  Root.  —  We  have  seen,  Art.  185,  that  all  pow 
ers  of  a  positive  quantity  are  positive  ;   but  the  even  powers 
of  a  negative  quantity  are  positive,  while  the  odd  powers  are 
negative. 

Thus  -fa,  when  raised  to  different  powers  in  succession, 
will  give  +ttj  _|_a2?  _|_a3?  +a4?  +a5?  _j_a6j  +a7^  etc> 

and  —  a,  in  like  manner,  will  give 

—a,  +  a2,  —a3,  -fa4,  —a5,  -fa6,  —a7,  etc. 

Hence  it  appears  that  if  the  root  to  be  extracted  be  express 
ed  by  an  odd  number,  the  sign  of  the  root  will  be  the  same  as 
the  sign  of  the  proposed  quantity.  Thus,  \/—  a3=  —a;  and 


If  the  root  to  be  extracted  be  expressed  by  an  even  number, 
and  the  quantity  proposed  be  positive,  the  root  may  be  either 
positive  or  negative.  Thus,  -\/~a?=  ±a. 

If  the  root  proposed  to  be  extracted  be  expressed  by  an  even 
number,  and  the  sign  of  the  proposed  quantity  be  negative,  the 
root  can  not  be  extracted,  because  no  quantity  raised  to  an 
even  power  can  produce  a  negative  result. 

196.  Hence,  to  extract  any  r#ot  of  a  monomial,  we  have  the 
following 

RULE. 

1st.  Extract  the  required  root  of  the  numerical  coefficient. 
Id.  Divide  the  exponent  of  each  literal  factor  ly  the  index  of  the 
required  root. 


EVOLUTION.  135 

3c7.  Every  even  root  of  a  positive  quantity  must  have  the  double 
sign  ±,  and  every  odd  root  of  any  quantity  must  have  the  same 
sign  as  that  quantity. 

From  Art.  186,  it  is  obvious  that  to  extract  any  root  of  a 
fraction,  we  must  divide  the  root  of  the  numerator  by  the  root 
of  the  denominator.  Thus, 

/ ^ 

a 

"I' 

EXAMPLES. 

1.  Find  the  square  root  of  64a6£4.  Am.  ±8a3&2. 

2.  Find  the  square  root  of  196a2&4c6x8.      Ans. 

3.  Find  the  square  root  of  225a2m68x6. 

4.  Find  the  cube  root  of  64a3&6x9.  Ans. 

5.  Find  the  cube  root  of  —  125a3x6?/9.  Ans.  —5ax2y3. 

6.  Find  the  cube  root  of  -343a6W. 

7.  Find  the  fourth  root  of  81a4&8.  Ans. 

8.  Find  the  fourth  root  of  256a4612x16. 

9.  Find  the  fifth  root  of  -32a56]0x15.  Ans.  - 


10.  Find  the  square  root  of  — — g— .  Ans.  ±- — — . 


11.  Find  the  square  root  of 


97/737,6 
12.  Find  the  cube  root  of  ~ 


2mx3' 


13.  Find  the  cube  root  of  — --_ 


216dV  * 

14.  Find  the  fourth  root  of         8  16. 

15.  Find  the  square  root  of  64a~2&~4x4.     Ans.  ±8a~lb~z 

16.  Find  the  cube  root  of  —  5l2a~3b~6x3. 

17.  Find  the  fourth  root  of  256a~4Z>-8x4. 

18.  Find  the  fifth  root  of  -32a~wb~l5x5.     Ans.  -2a~2b- 

19.  Find  the  square  root  of  (a  —  l>fx*.  Ans.  ±(a 

20.  Find  the  cube  root  of  (a+b)3(x+y)6. 

*b 


136  ALGEBKA. 

Square  Root  of  Polynomials. 

197.  In  order  to  discover  a  rule  for  extracting  the  square 
root  of  a  polynomial,  let  us  consider  the  square  of  a+&,  which 
is  a2-f-2a&  +  62,  If  we  arrange  the  terms  of  the  square  accord 
ing  to  the  dimensions  of  one  letter,  a,  the  first  term  will  be  the 
square  of  the  first  term  of  the  root;  and  since,  in  the  present 
case,  the  first  term  of  the  square  is  a2,  the  first  term  of  the  root 
must  be  a. 

Having  found  the  first  term  of  the  root,  we  must  consider 
the  rest  of  the  square,  namely,  2a&  +  62,  to  see  how  we  can  de 
rive  from  it  the  second  term  of  the  root.  Now  this  remainder 
may  be  put  under  the  form  (2a  +  b)b;  whence  it  appears  that 
we  shall  find  the  second  term  of  the  root  if  we  divide  the  re 
mainder  by  2a  +  &.  The  first  part  of  this  divisor,  2a,  is  double 
of  the  first  term  already  determined  ;  the  second  part,  &,  is  yet 
unknown,  and  it  is  necessary  at  present  to  leave  its  place  empty. 
Nevertheless,  we  may  commence  the  division,  employing  only 
the  term  2ct;  but  as  soon  as  the  quotient  is  found,  which  in 
the  present  case  is  6,  we  must  put  it  in  the  vacant  place,  and 
thus  render  the  divisor  complete. 

The  whole  process,  therefore,  may  be  represented  as  fol 
lows: 


If  the  square  contained  additional  terms,  we  might  continue 
the  process  in  a  similar  manner.  We  may  represent  the  first 
two  terms  of  the  root,  a  +5,  by  a  single  letter,  m,  and  the  re 
maining  terms  by  c.  The  square  of  m+c  will  be  mz+2mc+c2. 
The  square  of  the  first  two  terms  has  already  been  subtracted 
from  the  given  polynomial.  If  we  divide  the  remainder  by 
2m  as  a  partial  divisor,  we  shall  obtain  c,  which  we  place  in 
the  root,  and  also  at  the  right  of  2m,  to  complete  the  divisor. 
"We  then  multiply  the  complete  divisor  by  c,  and  subtract  the 


EVOLUTION.  137 

product  from  the  dividend,  and  thus  we  continue  until  all  the 
terms  of  the  root  have  been  obtained. 

198.  Hence  we  derive  the  following 

KULE. 

1st.  Arrange  the  terms  according  to  the  powers  of  some  one  let 
ter;  take  the  square  root  of  the  first  term  for  the  first  term  of  the 
required  root,  and  subtract  its  square  from  the  given  polynomial 

2d.  Divide  the  first  term  of  the  remainder  by  twice  the  root  al 
ready  found,  and  annex  the  result  both  to  the  root  and  the  divisor. 
Multiply  the  divisor  thus  completed  by  the  last  term  of  the  root,  and 
subtract  the  product  from  the  last  remainder. 

3d.  Double  the  entire  root  already  found  for  a  second  divisor. 
Divide  the  first  term  of  the  last  remainder  by  the  first  term  of  the 
second  divisor  for  the  third  term  of  the  root,  and  annex  the  result 
both  to  the  root  and  to  the  second  divisor,  and  proceed  as  before 
until  all  the  terms  of  the  root  have  been  obtained. 

If  the  given  polynomial  be  an  exact  square,  we  shall  at  last 
find  a  remainder  equal  to  zero. 

EXAMPLES. 

1.  Extract  the  square  root  of  a4— 2a3x+3a2x2— 2a 
a4  -  2a3x + BaV  -  2«x3 + x4  (  a2  -  ax + x* 


2a2— 

2a2x2— 


For  verification,  multiply  the  root  a2—  ax+x2  by  itself,  and 
we  shall  obtain  the  original  polynomial. 

2.  Extract  the  square  root  of  a24-2«&  +  2ac-f&2-f  2£c+c2. 

3.  Extract  the  square  root  of 


4  Extract  the  square  root  of 

8ax*  4-  4a2x2  +  4x4  +  1  662x2  -f  1  664  +  1  60  b2x. 

Ans 


138  ALGEBRA. 

5.  Extract  the  square  root  of 


6.  Extract  the  square  root  of  8a&3  +  a4—  4a3Z>+4&4. 

7.  Extract  the  square  root  of  4x4-l-12x3  +  5x'2—  6as+l. 


8.  Extract  the  square  root  of 

4z4  -  I2ax3  4-  25a2x2  -  24a3x     1  6a4. 


5.  2x2— 
9.  Extract  the  square  root  of 

-f  49a2x2  -  24a3x  +  1  6a4. 


10.  Extract  the  square  root  of  x*—  x3  +  ?  +4#—  2  + 

4 


x 


2     x     2 
'  X  -+- 


y 


199.  TFAen  a  Trinomial  is  a  Perfect  Square.  —  The  square  of 
a-\-~b  is  a2  +  2afr  +  &2,  and  the  square  of  a  —  b  is  a2—  2ab-\-b2. 
Hence  the  square  root  of  a2±2ab  +  b2  is  a±b;  that  is,  a  trino 
mial  is  a  perfect  square  when  two  of  its  terms  are  squares,  and 
the  third  is  the  double  product  of  the  roots  of  these  squares. 

Whenever,  therefore,  we  meet  with  a  quantity  of  this  de 
scription,  we  may  'know  that  its  square  root  is  a  binomial  ;  and 
the  root  may  be  found  by  extracting  the  roots  of  the  two  terms 
which  are  complete  squares,  and  connecting  them  by  the  sigp 
of  the  other  term. 

EXAMPLES. 

1.  Find  the  square  root  of  4«2+12a/;4-9&2.     Ans. 

2.  Find  the  square  root  of  9a2—  2 

3.  Find  the  square  root  of  9a4— 

4.  Find  the  square  root  of  4a2-fl4«&-f  16&2,  if  possible. 

,.w 

No  algebraic  binomial  can  be  a  perfect  square,  for  the  square 
of  a  monomial  is  a  monomial,  and  the  square  of  a  binomial 
necessarily  consists  of  three  distinct  terms. 


EVOLUTION.  139 

Square  Root  of  Numbers. 

200.  The  preceding  rule  is  applicable  to  the  extraction  of 
the  square  root  of  numbers;    for  every  number  may  be  re 
garded  as  an  algebraic  polynomial,  or  as  composed  of  a  cer 
tain  number  of  units,  tens,  hundreds,  etc.     Thus 

529  is  equivalent  to  500  +  20  +  9; 
also,  841  "  800+40  +  1. 

If,  then,  841  is  the  square  of  a  number  composed  of  tens 
and  units,  it  must  contain  the  square  of  the  tens,  plus  twice  the 
product  of  the  tens  by  the  units,  plus  the  square  of  the  units.  But 
these  three  terms  are  blended  together  in  841,  and  hence  arises 
the  peculiar  difficulty  in  determining  its  root.  The  following 
principles  will,  however,  enable  us  to  separate  these  terms, 
and  thus  detect  the  root. 

201.  Is/.  For  every  two  figures  of  the  square  there  will  be  one 
figure  in  the  root,  and  also  one  for  any  odd  figure.     Thus 


the  square  of    1  is  1 

"  9   "  81 

99   "        98,01 

999  "  99,80,01 


the  square  of      1  is  1 

10  "  1,00 

100  "        1,00,00 

1000  "  1,00,00,00 


The  smallest  number  consisting  of  two  figures  is  10,  arid  its 
square  is  the  smallest  number  of  three  figures.  The  smallest 
number  of  three  figures  is  100,  and  its  square  is  the  smallest 
number  of  five  figures,  and  so  on.  Therefore  the  square  root 
of  every  number  composed  of  one  or  two  figures  will  contain 
one  figure ;  the  square  root  of  every  number  composed  of  three 
or  four  figures  will  contain  two  figures;  of  a  number  from  five 
to  six  figures  will  contain  three  figures,  and  so  on. 

Hence,  if  we  divide  the  number  into  periods  of  two  figures, 
commencing  at  the.  units'  place,  the  number  of  periods  will 
indicate  the  number  of  figures  in  the  square  root. 

202.  2d.  The  first  figure  of  the  root  will  be  the  square  root  of 
the  greatest  square  number  contained  in  the  first  period  on  the  lejl. 


140  ALGEBRA. 

For  the  square  of  tens  can  give  no  significant  figure  in  the 
first  right  hand  period,  the  square  of  hundreds  can  give  no  fig- 
ure  in  the  first  two  periods  on  the  right,  and  the  square  of 
the  highest  figure  in  the  root  can  give  no  figure  except  in  the 
first  period  on  the  left. 

Let  it  be  required  to  extract  the  square  root  of  5329. 

This  number  contains  two 

periods,  indicating  that  there  53,29  (70  +  3,  the  root. 

will  be   two    places   in   the  49  00 

root.      Let  a  +  b  denote  the     140  +  3)~T29 
root,  where  a  is  the  value  of  4  29 

the  figure  in  the  tens'  place, 

and  I  of  that  in  the  units'  place.  Then  a  must  be  the  great 
est  multiple  of  10,  which  has  its  square  less  than  5300 ;  this  is 
found  to  be  70.  Subtract  a2,  that  is  the  square  of  70,  from 
the  given  number,  and  the  remainder  is  429,  which  must  be 
equal  to  (2a  +  b)b.  Divide  this  remainder  by  2a,  that  is  by 
140,  and  the  quotient  is  3,  which  is  the  value  of  b.  Com 
pleting  the  divisor,  we  have  2a  +  6  =  143;  whence  (2a  +  b)b, 
that  is  143  X  3,  or  429,  is  the  quantity  to  be  subtracted ;  and 
as  there  is  now  no  remainder,  we  conclude  that  70  +  3,  or  73, 
is  the  required  square  root. 

For  the  sake  of  brevity,  the  ciphers  may  be  omitted,  pro 
vided  we  retain  the  proper  local  values  of  the  figures. 

If  the  root  consists  of  three  places  of  figures,  let  a  represent 
the  hundreds,  and  b  the  tens;  then,  having  obtained  a  and  b  as 
before,  let  the  hundreds  and  tens  together  be  considered  as  a 
new  value  of  a,  and  find  a  new  value  of  b  for  the  units. 

Required  the  square  root  of  568516. 

Having  found  75,  the  square  root  of  56,85,16(754 

che  greatest  square  number  contained  in  49 

the  first  two  periods,  we  bring  down  the      145 )  785 
last  period,  and  have  6016  for  a  new  div-  725 

idend.     We  then  take  2a,  or  150,  for  a     1504)  6016 
partial  divisor,  whence  we  obtain  b  =  4:  6016 

for  the  last  figure  of  the  root.     The  en 
tire  root  is  therefore  754. 


EVOLUTION.  141 

203.  Hence,  for  the  extraction  of  the  square  root  of  num 
bers,  we  derive  the  following 

RULE. 

1st.  Separate  the  given  number  into  periods  of  two  figures  each, 
beginning  from  the  units1  place. 

2d.  Find  the  greatest  number  whose  square  is  contained  in  the 
left-hand  period  ;  this  is  the  first  figure  of  the  required  root.  Sub 
tract  its  square  from  the  first  period,  and  to  the  remainder  bring 
down  the  second  period  for  a  dividend. 

3d.  Double  the  root  already  found  for  a  divisor ,  and  find  how 
many  times  it  is  contained  in  the  dividend,  exclusive  of  its  right- 
hand  figure ;  annex  the  result  both  to  the  root  and  the  divisor. 

4dh.  Multiply  the  divisor  thus  increased  by  the  last  figure  of  the 
root,  subtract  the  product  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend. 

6th.  Double  the  whole  root  now  found  for  a  new  divisor,  and 
proceed  as  before,  continuing  the  operation  until  all  the  periods  are 
brought  down. 

In  applying  the  preceding  rule,  it  may  happen  that  the  prod 
uct  of  the  complete  divisor  by  the  last  figure  of  the  root  is 
greater  than  the  dividend.  This  indicates  that  the  last  figure 
of  the  root  was  taken  too  large,  and  this  happens  because ^the 
divisor  is  at  first  incomplete — that  is,  is  too  small.  In  such  a 
case,  we  must  diminish  the  last  figure  of  the  root  by  unity  until 
we  obtain  a  product  which  is  not  greater  than  the  dividend. 

EXAMPLES. 

1.  What  is  the  square  root  of  294849?  Ans.  543, 

2.  What  is  the  square  root  of  840889  ? 

3.  What  is  the  square  root  of  1142761  ? 

4.  What  is  the  square  root  of  32239684? 

5.  What  is  the  square  root  of  72777961  ? 

6.  What  is  the  square  root  of  3518743761  ? 

204.  Square  Root  of  Fractions. — We  have   seen  that  the 
root  of  a  fraction  is  equal  to  the  root  of  its  numerator  di- 


142  ALGEBRA. 

vided  by  the  root  of  its  denominator     Hence  the  square  root 


The  number  5.29  may  be  written  —  -,  and  its  square  root 

O  D  -j  Qfif\f)  A 

is  —  ,  or  2.3.     So,  also,  18.6624  may  be  written  ,  and 

its  square  root  is  —  v  or  4.32.     That  is,  the  square  root  of  a 
100 

decimal  fraction,  or  of  a  whole  number  followed  by  a  decimal 
fraction,  may  be  found  in  the  same  manner  as  that  of  a  whole 
number,  if  we  divide  it  into  periods  commencing  with  the  deci 
mal  point. 

In  the  extraction  of  the-  square  root  of  an  integer,  if  there  is 
still  a  remainder  after  we  have  obtained  the  units'  figure  of  the 
root,  it  indicates  that  the  proposed  number  has  not  an  exact 
square  root.  We  may,  if  we  please,  proceed  with  the  approxi 
mation  to  any  desired  extent  by  supposing  a  decimal  point  at 
the  end  of  the  proposed  number,  and  annexing  any  even  num 
ber  of  ciphers,  and  continuing  the  operation.  We  thus  obtain 
a  decimal  part  to  be  added  to  the  integral  part  already  found. 

So,  also,  if  a  decimal  number  has  no  exact  square  root,  we 
may  annex  ciphers  and  proceed  with  the  approximation  to 
any  desired  extent. 

EXAMPLES. 

TX7,    ,  .     ,  -  1089  0  33 

1.  What  is  the  square  root  of  ?  Am.  —  . 

^ooy  uo 

2.  What  is  the  square  root  of 


-^ 
1/0569 

3.  What  is  the  square  root  of  „,—  -.,,.„.,  ? 

67551961 

4.  What  is  the  square  root  of  9.878449  ? 

5.  What  is  the  square  root  of  58.614336  ? 

6.  What  is  the  square  root  of  .558009  ? 

7.  What  is  the  square  root  of  .03478225  ? 

Find  the  square  roots  of  the  following  numbers  to  five  deci 
mal  places. 


EVOLUTION.  143 


8.  Of  2.         Ans.  1.41421. 

9.  Of  10.       Ans.  3.16227. 
10.  Of  9.1. 


11.  Of  4f. 

12.  Of  TV 

13.  Of  A. 


Cube  Root  of  a  Polynomial. 

205.  We  already  know  that  the  cube  of  a+b  is  a 
-\-3ab2  +  b3.     If,  then,  the  cube  were  given,  and  we  were   re 
quired  to  find  its  root,  it  might  be  done  by  the  following 
method. 

When  the  terms  are  arranged  according  to  the  powers  of 
one  letter,  a,  we  at  once  know,  from  the  first  term,  a3,  that 
a  must  be  one  term  of  the  root.  If,  then,  wre  subtract  its 
cube  from  the  proposed  polynomial,  we  obtain  the  remainder 
8a2b  +  Bab2  <f  63,  which  must  furnish  the  second  term  of  the  root. 

Now  this  remainder  may  be  put  under  the  form 


whence  it  appears  that  we  shall  find  the  second  term  of  the 
root  if  we  divide  the  remainder  by  3a2+3ab  +  b2.  But,  as  this 
second  term  is  supposed  to  be  unknown,  the  divisor  can  not  be 
completed.  Nevertheless,  we  know  the  first  term,  3a2,  that  is 
thrice  the  square  of  the  first  term  already  found,  and  by  means 
of  this  we  can  find  the  other  part,  b;  viz.,  by  dividing  the  first 
term  of  the  remainder  by  3a2.  We  then  complete  the  divisor 
by  adding  to  it  Sab+b2,  If  this  complete  divisor  be  multi 
plied  by  6,  it  will  give  the  last  three  terms  of  the  power. 
Let  it  be  required  to  find  the  cube  root  of  8a3 


Having  found  the  first  term  of  the  root,  2a,  and  subtracted 
its  cube,  we  divide  the  first  term  of  the  remainder,  36«25,  by 
three  times  the  square  of  2a,  that  is  12a2,  and  we  obtain  Sb 
for  the  second  term  of  the  root.  We  then  complete  the  divi 
sor  by  adding  to  it  three  times  the  product  of  the  two  terms  of 


144  ALGEBKA. 

the  root,  which  is  18a5,  together  with  the  square  of  the  last 
term  36,  which  is  9&2.     Multiplying  then  the  complete  divisoi 
by  3Z>,  and  subtracting  the  product  from  the  last  remainder, 
nothing  is  left.     Hence  the  required  cube  root  is  2a+35. 
This  result  may  be  easily  verified  by  multiplication. 


206.  If  the  root  contains  three  terms,  as  a  +  6  +  c,  we  may 
put  a  +  b  =  m.     Then 

(a  +  b  +  c)3  =  (m  +  c)3  =  m3  +  3ra2c  +  3mc2  +  c3. 
If  we  proceed  as  in  the  last  example,  we  shall  find  a  +  &,  and 
we  subtract  its  cube  from  the  given  polynomial.     There  will 
then  remain  3m2c+3?wc2  +  c3,  which  may  be  written 

(3ra2  +  3rac+c2)c. 

We  perceive  that  3m2  will  be  the  new  trial  divisor  to  obtain 
c.     We  then  complete  the  divisor  by  adding  to  it  Smc-fc2. 
Let  it  be  required  to  find  the  cube  root  of  8a6  —  365a54- 
a*  -  63£3a3 


8a6  -  366rt5  +  GGZ*2a4  -  63Z>3a3  +  336*a2  -  966o  +  66(2a2  -  3ba  +  63 
So6 


1 2a*  —  1 8ba3  +  9/>2a2  )  —  366a5  +  GG6~a*  —  G363«3 

—  366a5  +  5462a4  —  2  7i3a3 

12a*- 


The  first  term  of  the  root  is  2a2,  and  subtracting  its  cube,  the 
first  term  of  the  remainder  is  —  365a5,  which,  divided  by  3 
times  the  square  of  2a2,  gives  —  Sba  for  the  second  term  of  the 
root.  Complete  the  divisor  as  in  the  last  example,  and  multi 
ply  it  by  —3ba.  Subtracting  the  product  from  the  last  remain 
der,  the  first  term  of  the  second  remainder  is  1262a4. 

To  form  the  new  trial  divisor,  we  take  three  times  the  square 
of  the  part  of  the  root  already  found,  viz.,  2cr2— 3ba.  Divide 
the  first  term  of  the  remainder  by  12a*,  and  we  obtain  bz  for  the 
last  term  of  the  root.  We  now  complete  the  divisor  by  add 
ing  to  it  three  times  the  product  of  the  third  term  by  the  sum 
of  the  first  two  terms,  and  also  the  square  of  the  last  term, 
Multiplying  the  divisor  thus  completed  by  52,  we  find  the  prod- 


EVOLUTION.  145 

uct  equal  to  the  last  remainder.     Hence  the  required  cube  root 
is2a2— 


207.  Hence,  for  extracting  the  cube  root  of  a  polynomial,  we 
derive  the  following 

RULE. 

1st.  Arrange  the  terms  according  to  the  powers  of  some  one  let 
ter  ;  take  the  cube  root  of  the  first  term,  and  subtract  the  cube  from 
the  given  polynomial. 

2d.  Divide  the  first  term  of  the  remainder  by  three  times  the 
square  of  the  root  already  found  ;  the  quotient  will  be  the  second 
term  of  the  root. 

3d.  Complete  the  divisor  by  adding  to  it  three  times  the  product 
of  the  two  terms  of  the  root  and  the  square  of  the  second  term. 

4th.  Multiply  the  divisor  thus  increased  by  the  last  term  of  the 
root,  and  subtract  the  product  from  the  last  remainder. 

5th.  Take  *  three  times  the  square  of  the  part  of  the  root  already 
found  for  a  new  trial  divisor,  and  proceed  by  division  to  find  an 
other  term  of  the  root. 

6th.  Complete  the  divisor  by  adding  to  it  three-  times  the  product 
of  the  last  term  by  the  sum  of  the  first  two  terms,  and  also  the  square 
of  the  last  term,  with  which  proceed  as  before  till  the  entire  root  has 
been  obtained. 

We  may  dispense  with  forming  the  complete  divisor  accord 
ing  to  the  rule  if  each  time  that  we  find  a  new  term  of  the 
root  we  raise  the  entire  root  already  found  to  the  third  power, 
and  subtract  the  cube  from  the  given  polynomial. 

EXAMPLES. 

1.  What  is  the  cube  root  of  a6—  6a5  +  15a4—  20a3  +  15a2— 
8a-f  1?  Ans.  a2- 

2.  What  is  the  cube  root  of  6x6-40x3+z6  +  96;r 

Ans. 

3.  What  is  the  cube  root  of  18x4  +  36x2-f  24cc+  8  +  32x3-f- 
x6  +  6xfl?  Ans. 

4.  What  is  the  cube  root  of  3£6  +  &6  -  563  -  1  +  36  ? 

Ans. 
G 


146  ALGEBRA. 

5.  What  is  the  cube  root  of  8x6—  36cc5  +  66x4—  63x3-h33x2-^ 
9x+l? 

6.  What  is  the  cube  root  of  8x6+4;8ax5+  60a2x4—  80a3z3— 


7.  What  is  the  cube  root  of  8x6  —  36ax5  +  102a2x4—  171a3a3 
5z  +  64a6  ? 


./foctf  o/"  Numbers. 

208.  The  preceding  rule  is  applicable  to  the  extraction  of 
the  cube  root  of  numbers  ;  but  a  difficulty  in  applying  it  arises 
from  the  fact  that  the  terms  of  the  powers  are  all  blended  to 
gether  in  the  given  number.  They  may,  however,  be  separated 
by  attending  to  the  following  principles  : 

1st.  For  every  three  figures  of  the  cube  there  will  be  one  figure  in 
the  root,  and  also  one  for  any  additional  figure  or  figures.  Thus, 


the  cube  of    1  is  1 

9  "  729 

99  "         970,299 
999  "  997,002,999 


the  cube  of      1  is  1 

10  "  1,000 

"         100  "         1,000,000 

1000  "  1,000,000,000 


Hence  we  see  that  the  cube  root  of  a  number  consisting  of 
from  one  to  three  figures  will  contain  one  figure ;  the  cube  root 
of  a  number  consisting  of  from  four  to  six  figures  will  contain 
two  figures ;  of  a  number  from  seven  to  nine  figures  will  con 
tain  three  figures,  and  so  on. 

Hence,  if  we  divide  the  number  into  periods  of  three  figures, 
commencing  at  units'  place,  the  number  of  periods  will  indi 
cate  the  number  of  figures  in  the  cube  root 

209.  2c?.  The  first  figure  of  the  root  will  be  the  cube  root  of  the 
greatest  cube  number  contained  in  the  first  period  on  the  left. 

For  the  cube  of  tens  can  give  no  significant  figure  in  the  first 
right-hand  period  ;  the  cube  of  hundreds  can  give  no  figure  in 
the  first  two  periods  on  the  right ;  and  the  cube  of  the  high 
est  figure  in  the  root  can  give  no  figures  except  in  the  first 
period  on  the  left. 

Let  it  be  required  to  extract  the  cube  root  of  438976. 

This  number  contains  two  periods,  indicating  that  there  will 


EVOLUTION.  147 


be  two  places  in  438,976  ( 70  f  6,  the  root, 

the  root.    Let  a  be  343,000 

f    ,,  70ax3  =  H700 

the   value    of  the          70x6x3-  1260 


95976 
95976 


figure  in  the  tens'  G2— 5? 

place,  and  b  of  that  comPlete  divisor> 15996 
in  the  units'  place.  Then  a  must  be  the  greatest  multiple  of 
10  which  has  its  cube  less  than  438000 ;  that  is,  a  must  be  70. 
Subtract  the  cube  of  70  from  the  given  number,  and  the  re 
mainder  is  95976.  This  remainder  corresponds  to  3a2b  -\-3ab* 
+  &3,  which  may  be  written 


Divide  this  remainder  by  3a2,  that  is,  by  14700,  and  the  quo 
tient  is  6,  which  is  the  value  of  b.  Complete  the  divisor  by 
adding  to  it  3a5,  or  1260,  and  b2,  or  36.  The  complete  divisor 
is  thus  found  to  be  15996,  which,  multiplied  by  6,  gives  95976. 
Subtracting,  the  remainder  is  zero,  and  we  conclude  that  70  +  6, 
or  76,  is  the  required  cube  root. 

For  the  sake  of  brevity  the  ciphers  may  be  omitted,  provided 
we  retain  the  proper  local  values  of  the  figures. 

If  the  root  consists  of  more  than  two  places  of  figures,  the 
method  will  be  substantially  the  same. 

Let  it  be  required  to  extract  the  cube  root  of  279,726,264. 
O^Q  7o«  OA/I  fAKA         Having  found  65,  the  cube  root  of 

27y,7ZO,4O4:(o54:       Xl  ,  ,   .         ,        „ 

the  greatest  cube  contained  in  the  first 
two  periods,  we  bring  down  the  last 
period,  and  have  5101264  for  a  new 
dividend.  We  then  take  three  times 
the  square  of  the  root  already  found, 
5101  264  or  12675,  for  a  partial  divisor,  whence 

5101  264  we  °ktam  4  for  the  last  figure  of  the 

—  root.     We  then  complete  the  divisor 

by  adding  to  it  three  times  the  product  of  4  by  65,  and  the 
square  of  4,  regard  being  paid  to  the  proper  local  values  of  the 
figures.  The  complete  divisor  is  thus  found  to  be  1275316, 
which,  multiplied  by  4,  gives  5101264.  Hence  654  is  the  re« 
quired  cube  root. 


108 


90 


11725 
12675 

780 

__  16 
1275316 


148  ALGEBEA. 

210.  Hence,  for  the  extraction  of  the  cube  root  of  numbers, 
we  derive  the  following 

RULE. 

1st.  Separate  the  given  number  into  periods  of  three  figures  each, 
"beginning  at  the  units'  place. 

2d.  Find  the  greatest  cube  contained  in  the  left-hand  period ;  its 
cube  root  is  the  first  figure  of  the  required  root.  Subtract  the  cube 
from  the  first  period,  and  to  the  remainder  bring  down  the  second 
period  for  a  dividend. 

3d.  Take  three  hundred  times  the  square  of  the  root  already  found 
for  a  trial  divisor;  find  how  many  times  it  is  contained  in  the 
dividend,  and  write  the  quotient  for  the  second  figure  of  the  root. 

4th.  Complete  the  divisor  by  adding  to  it  thirty  times  the  product 
of  the  two  figures  of  the  root,  and  the  square  of  the  second  figure. 

6th.  Multiply  the  divisor  thus  increased  by  the  last  figure  of  Hie 
root;  subtract  the  product  from  the  dividend,  and  to  the  remainder 
"bring  down  the  next  period  for  a  new  dividend. 

6th.  Take  three  hundred  times  the  square  of  the  whole  root  now 
found  for  a  new  trial  divisor,  and  by  division  obtain  another  figure 
of  the  root. 

1th.  Complete  the  divisor  by  adding  to  it  thirty  times  the  product 
of  the  last  figure  by  the  former  figures,  and  also  the  square  of  the 
last  figure,  with  which  proceed  as  before,  continuing  the  operation 
until  all  the  periods  are  brought  down. 

It  will  be  observed  that  three  times  the  square  of  the  tens, 
when  their  local  value  is  regarded,  is  the  same  as  three  hund 
red  times  the  square  of  this  digit,  not  regarding  its  local  value. 

In  applying  the  preceding  rule,  it  may  happen  that  the  prod 
uct  of  the  complete  divisor  by  the  last  figure  of  the  root  is 
greater  than  the  dividend.  This  indicates  that  the  last  figure 
of  the  root  was  taken  too  large,  and  this  happens  because  the 
divisor  is  at  first  incomplete,  that  is,  too  small.  In  such  a  case 
we  must  diminish  the  last  figure  of  the  root  by  unity,  until  we 
obtain  a  product  which  is  not  greater  than  the  dividend. 

EXAMPLES. 

1.  Find  the  cube  root  of  163667323.  Ans.  547. 


EVOLUTION.  149 

2.  Find  the  cube  root  of  39651821.  Ans  341. 

3.  Find  the  cube  root  of  4019679.  Ans.  159. 

4.  Find  the  cube  root  of  12895213625. 

5.  Find  the  cube  root  of  183056925752. 

6.  Find  the  cube  root  of  759299343867. 

211.  Cube  Root  of  Fractions. — The  cube  root  of  a  fraction  is 
equal  to  the  root  of  its  numerator  divided  by  the  root  of  its  de 
nominator.  Hence  the  cube  root  of  -YTT$  is  T7^. 

The  number  12.167  may  be  written  \Voir-5  and  its  cube  root 
is  f-f,  or  2.3.  That  is,  the  cube  root  of  a  decimal  fraction,  or 
of  a  whole  number  followed  by  a  decimal  fraction,  may  be 
found  in  the  same  manner  as  that  of  a  whole  number,  if  we 
divide  it  into  periods  commencing  with  the  decimal  point. 

In  the  extraction  of  the  cube  root  of  an  integer,  if  there  is 
still  a  remainder  after  we  have  obtained  the  units'  figure  of  the 
root,  it  indicates  that  the  proposed  number  has  not  an  exact 
cube  root.  We  may,  if  we  please,  proceed  with  the  approxi 
mation  to  any  desired  extent,  by  supposing  a  decimal  point  at 
the  end  of  the  proposed  number,  and  annexing  any  number 
of  periods  of  three  ciphers  each,  and  continuing  the  operation. 
We  thus  obtain  a  decimal  part  to  be  added  to  the  integral  part 
already  found. 

So,  also,  if  a  decimal  number  has  no  exact  cube  root,  we  may 
annex  ciphers,  and  proceed  with  the  approximation  to  any  de 
sired  extent. 

EXAMPLES. 

1.  Find  the  cube  root  of  A8^-.  Ans.  £f . 

2.  Find  the  cube  root  of  14^-.  Ans.  2f 

3.  Find  the  cube  root  of  13.312053. 

4.  Find  the  cube  root  of  1892.819053. 

5.  Find  the  cube  root  of  .001879080904. 

Find  the  cube  roots  of  the  following  numbers  to  5  decimal 
places : 


6.  15.25. 

Ans.  2.47984. 

10.  11. 

7.  3.7. 

Ans.  1.54668. 

11.  f 

8.  100.1. 

Ans.  4.64314. 

12.  f. 

9.  4. 

Ans.  1.58740. 

150  ALGEBRA. 


CHAPTER  XIII. 

RADICAL    QUANTITIES. 

212.  A  radical  quantity  is  an  indicated  root  of  a  quantity: 
as  ^/a,  y/a,  etc.     Radical  quantities  may  be  either  surd  or  ra 
tional. 

Radical  quantities  are  divided  into  degrees,  the  degree  being 
denoted  by  the  index  of  the  root.  Thus,  V3  is  a  radical  of  the 
second  degree ;  v/5  is  a  radical  of  the  third  degree,  etc. 

213.  The  coefficient  of  a  radical  is  the  number  or  letter  pre 
fixed  to  it,  showing  how  often  the  radical  is  to  be  taken.    Thus, 
in  the  expression  2  -v/^  2  is  the  coefficient  of  the  radical. 

Similar  radicals  are  those  which  have  the  same  index  and 
the  same  quantity  under  the  radical  sign.  Thus,  B\/a  and  5\/a 
are  similar  radicals.  Also  7Vb  and  10v/5  are  similar  radicals. 

214.  Use  of  fractional  Exponents. — We  have  seen,  Art.  196, 
that  in  order  to  extract  any  root  of  a  monomial,  we  must  di 
vide  the  exponent  of  each  literal  factor  by  the  index  of  the  re 
quired  root.     Thus  the  square  root  of  a4  is  a2,  and  in  the  same 

manner  the  square  root  of  a3  may  be  written  a*,  that  of  a5  will 

5  1 

be  a*,  and  that  of  a,  or  a1,  is  a  .     Whence  we  see  that 
a    is  equivalent  to  Va, 
a?  "  Vo5, 

a*  "  Ve?,  etc. 

2 

So,  also,  the  cube  root  of  a2  may  be  written  a   ;  the  cube 

4  1 

root  of  a4  is  a7;  and  the  cube  root  of  a,  or  a1,  is  a  .     Whence 
we  see  that 


RADICAL  QUANTITIES. 

a    is  equivalent  to  V/a, 
a*  "  I/a*, 

a?  "  V/a",  etc. 

In  the  same  manner,  or  is  equivalent  to  v/a» 
a^  "  Va, 

a?  "  I/a5, 


a"  KU/  . 

That  isj  $e  numerator  of  a  fractional  exponent  denotes  the  power, 
and  the  denominator  the  root  to  be  extracted. 

Let  it  be  required  to  extract  the  cube  root  of  — -.  This  quan 
tity,  Art.  187,  is  equivalent  to  a~4.  Now,  to  extract  the  cube 
root  of  a~\  we  must  divide  its  exponent  by  3,  which  gives  us 

-4  1  1 

a  T.     But  the  cube  root  of  -j  may  also  be  represented  by  — 


Hence  —  is  equivalent  to  a  T. 

a* 

So,  also,  -T  is  equivalent  to  a     , 

a 

4      «      .-«; 

a~ 
1  _= 


Thus  we  see  that  the  principle  of  Art.  77,  that  a  factor  may 
be  transferred  from  the  numerator  to  the  denominator  of  a  frac 
tion,  or  from  the  denominator  to  the  numerator  by  changing 
the  sign  of  its  exponent,  is  applicable  also  to  fractional  exponents. 

"We  may  therefore  entirely  reject  the  radical  signs  hitherto 
employed,  and  substitute  for  them  fractional  exponents,  and 
many  of  the  difficulties  which  occur  in  the  reduction  of  radical 
quantities  are  thus  made  to  disappear. 


152  ALGEBRA. 

To  reduce  a  Radical  to  its  simplest  Form. 

215.  A  radical  is  in  its  simplest  form  when  it  has  under  the 
radical  sign  no  factor  which  is  a  perfect  power  corresponding 
to  the  degree  of  the  radical. 

Radical  quantities  may  frequently  be  simplified  by  the  ap 
plication  of  the  following  principle  :  the  nth  root  of  the  product 
of  two  or  more  factors  is  equal  to  the  product  of  the  nth  roots  of  those 
factors  ;  or,  in  algebraic  language, 


For  each  of  these  expressions,  raised  to  the  nth  power,  will 
give  the  same  quantity. 

Thus,  the  nth  power  of  V~ab  is  ah. 
And  the  nib.  power  of  Va  X  VI  is  (Va)n  x  (\fbT,  or  db. 

Hence,  since  the  same  powers  of  the  quantities  n]/ab  and 
v/a  X  Vb  are  equal,  the  quantities  themselves  must  be  equal. 

Let  it  be  required  to  reduce  V48a3x2  to  its  simplest  form. 

This  expression  may  be  put  under  the  form  Vl6a2#2  x  VSa. 

But  Vl6a2x2  is  equal  to  4ax. 

Hence  V^Sa3x2=4ax  V3a. 

Hence,  to  reduce  a  radical  to  its  simplest  form,  we  have  the 
following 

EULE. 

Resolve  the  quantity  under  the  radical  sign  into  two  factors,  one 
of  which  is  the  greatest  perfect  power  corresponding  in  degree  to  the 
radical.  Extract  the  required  root  of  this  factor,  and  prefix  it  to 
the  other  factor,  which  must  be  left  under  the  sign. 

EXAMPLES. 

Eeduce  the  following  radicals  to  their  simplest  forms. 

l25a3.  Ans.  5a  V5a. 

.  W\/2aa 


KADICAL   QUANTITIES. 


153 


6. 


216.  When  the  quantity  under  the  radical  sign  is  a  fraction, 
it  is  often  convenient  to  multiply  both  its  terms  by  such  a 
quantity  as  will  make  the  denominator  a  perfect  power  of  the 
degree  indicated.  Then,  after  simplifying,  the  factor  remain 
ing  under  fohe  radical  sign  will  be  entire. 

14.  3  A/|.  Ans.  SVI^SVrl^lVlO. 

15.  4-Y/1+6V3?.  Ans.  |V2-f3VT4. 


Ans. 


IVlo. 


.  2  Vab. 


Ans.  j- 
b 


23.  23v/i+3v/i. 
24. 


Ans. 


Ans. 


217.  The  following  principle  can  frequently  be  employed  in 
simplifying  radicals : 

TJie  mnih  root  of  any  quantity  is  equal  to  the  mth  root  of  the  nth 

root  of  that  quantity.     That  is, 

G  2 


154:  ALGEBRA. 


For,  if  we  raise  each  expression  to  the  mth  power,  it  becomes  V#. 
Thus,  the  fourth  root = the  square  root  of  the  square  root. 

the  sixth  root  =  the  square  root  of  the  cube  root,  or 

the  cube  root  of  the  square  root, 
the  eighth  root = the  square  root  of  the  fourth  root,  or 
the  fourth  root  of  the  square  root, 
the  ninth  root  =the  cube  root  of  the  cube  root. 
Hence,  ivhen  the  index  of  a  root  is  the  product  of  two  or  more 
factors,  we  may  obtain  the  root  required  by  extracting  in  succession 
the  roots  denoted  by  those  factors. 

Ex.  1.  Let  it  be  required  to  extract  the  sixth  root  of  64. 
The  square  root  of  64  is  8,  and  the  cube  root  of  8  is  2.    Hence 
the  sixth  root  of  64  is  2. 

Ex.  2.  Extract  the  eighth  root  of  256.  Ans.  2. 

Ex.  3.  Find  the  fourth  root  of  1874161.  Ans.  37. 

Ex.  4.  Find  the  sixth  root  of  148035889.  Ans.  23. 

Ex.  5.  Find  the  ninth  root  of  387420489.  Ans.  9. 

Ex.  6.  Find  the  eighth  root  of  2562890625.  Ans.  15. 

218.  When  the  index  of  a  root  is  the  product  of  two  or  more 
factors,  and  one  of  the  roots  can  be  extracted,  while  the  other 
can  not,  a  radical  may  be  simplified  by  extracting  one  of  the  roots. 

Thus,  t/9=\/3. 

Keduce  the  following  radicals  to  their  simplest  forms : 

Ex.  1.  6v/4o"2.  Ans.  ]/fa 

Ex.2. 

Ex.3. 

Ex.4. 


Ex.5. 

Ex.6.  \/^4-.  Ans. 


RADICAL   QUANTITIES.  155 

To  introduce  a  Factor  under  the  Radical  Sign. 

219.  The  square  root  of  the  square  of  a  is  obviously  a,  and 
the  cube  root  of  the  cube  of  a  is  a,  etc.     That  is, 

a=}/cP  =  Va?  =  Vdi1  etc. 
Whence,  also,  a  \/l  —  v/o5  X  \/b  =  \/a2b. 

Hence,  to  introduce  a  factor  under  the  radical  sign,  we  have 
the  following 

RULE. 

Raise  the  factor  to  a  power  denoted  by  the  index  of  the  required 
root,  and  write  it  as  a  factor  under  the  radical  sign. 

EXAMPLES. 

1.  Keduce  ax2  to  a  radical  of  the  second  degree. 

%  Ans. 

2.  Eeduce  2azbx  to  a  radical  of  the  third  degree. 

Ans. 

3.  Keduce  5  +  6  to  a  radical  of  the  second  degree. 

Ans. 


Transform  the  following  radicals  by  introducing  the  coefrV 
cients  as  factors  under  the  radical  sign  : 

4.  6V8J.  Ans.  Vl26. 

5.  4</J+3iV§.  Ans.  -V/2+V98. 


6.  a-+a&-iT-  Ans. 

ab 


8.  2V2  +  71/5. 

9.  r|v/7j|. 

^To  change  the  Index  of  a  Radical. 
220.  From  Art.  219,  it  follows  that 

Va  =  W2  =  V^3  =  v/o*,  etc. ; 
V/a  =  v/a5  =  I/a5  =  Va*,  etc. ; 
V^=v/a"2  =  Va1  =  {X?,  etc. 


156  ALGEBRA. 

Hence  we  see  that  the  index  of  any  radical  may  be  multiplied 
by  any  number,  provided  we  raise  the  quantity  under  the  radical 
sign  to  a  power  whose  exponent  is  the  same  number  ;  or  the  index 
of  any  radical  may  be  divided  by  any  number,  provided  we  extract 
that  root  of  the  quantity  under  the  radical  sign  whose  index  is  the 
same  number. 

If,  instead  of  the  radical  sign,  we  employ  fractional  expo 
nents,  we  shall  have 

1234 
1234 

a*=cP=ci*=aT*,  etc. 

Hence  we  see  that  we  may  multiply  or  divide  both  terms  of  a 
fractional  exponent  by  the  same  number,  without  changing  the 
value  of  the  expression. 

EXAMPLES. 

Verify  the  following  equations : 

1.  V2+VS-fVi=Vi-hV5+Vl0. 

2.  t/^Ts. 


8.    2Va*. 

4.  2  V^-b  =  2  Va3 -3azb+ 3a£2 - b3. 

5.  SVaic 


To  reduce  Radicals  to  a  Common  Index. 

221.  Let  it  be  required  to  reduce  Va  and  Va  to  equivalent 
radicals  having  a  common  index.  Substituting  for  the  radical 
signs  fractional  exponents,  the  given  quantities  are 


a?  and  a* 


Keducing  the  exponents  to  a  common  denominator,  the  ex 
pressions  are  a  2 

a*  and  OT, 

or  Va?  and  Va\ 


RADICAL   QUANTITIES.  157 

which  are  of  the  same  value  as  the  given  quantities,  and  have 
a  common  index  6.     Hence  we  derive  the  following 

RULE. 

Reduce  the  fractional  exponents  to  a  common  denominator,  raise, 
each  quantity  to  the  power  denoted  ly  the  numerator  of  its  new  ex 
ponent,  and  take  the  root  denoted  ~by  the  common  denominator. 

EXAMPLES. 
11  1 

1.  Reduce  a  ,  a  ,  and  a?  to  a  common  index. 


3 

TT 


Ans.  a1"*,  a^,  and  a 

1  2 

2.  Eeduce  a?,  a2,  and  b*  to  a  common  index. 

3          12  4 

Ans.  a  ,  a    ,  and  8*. 
11  i 

3.  Reduce  2T,  3T,  and  5T  to  a  common  index. 

Ans.  V6i,  "v/81,  and  'v/125. 

12  3 

4.  Reduce  3T,  2T,  and  2T  to  a  common  index. 

.  12v/729,  "1/256,  and  'v/512. 


6.  Reduce  o  ,  am,  and  an  to  a  common  index. 

mn          2»i  2m 

J.W5.  a2771",  a2nm,  and  a2""*. 

6.  Reduce  y/3,  VB,  and  v/7  to  a  common  index. 

7.  Reduce  V2ab,  VSab2,  and  V&ab3  to  a  common  index. 

8.  Reduce  Va+6,  V«—  ^  ^n(l  V«2—  &2  to  a  common  index. 

To  add  Radical  Quantities  together. 

222.  When  the  radical  quantities  are  similar,  the  common 
radical  part  may  be  regarded  as  the  unit,  and  the  coefficient 
shows  how  many  times  this  unit  is  repeated.  The  sum  of  the 
coefficients  of  the  given  radicals  will  then  denote  how  many 
times  this  unit  is  to  be  repeated  in  the  required  sum. 

If  the  radicals  are  not  similar  they  can  not  be  added,  because 
they  have  no  common  unit  In  such  a  case,  the  addition  can 


158  ALGEBKA. 

only  be  indicated  by  the  algebraic  sign.  Eadicals  which  are 
apparently  dissimilar  may  become  similar  when  reduced  to 
their  simplest  forms.  Hence  we  have  the  following 

RULE. 

Reduce  each  radical  to  its  simplest  form.  If  the  resulting  radi 
cals  are  similar,  add  their  coefficients,  and  to  their  sum  annex  the 
common  radical.  If  they  are  dissimilar ,  connect  them  by  the  sign 
of  addition. 

EXAMPLES. 

1.  Find  the  sum  of  V27,  V58,  and  V75.         Ans.  12  VS. 

2.  Find  the  sum  of  4Vt±7r  3  A/75,  and  Vl92. 

Ans.  51  VS. 

3.  Find  the  sum  of  V72,  VI28,  and  V162.     Ans.  23  V2. 

4.  Find  the  sum  of  A/180,  A/405,  and  A/320.    Ans.  23  A/5. 

5.  Find  the  sum  of  3  A/|,  2  V^V,  and  4A/S-      ^s-  VI5. 

6.  Find  the  sum  of  'v/500,  V/108,  and  \/256.     ^w&  12  V3. 

7.  Find  the  sum  of  V50,  1/135,  and  v/320.        ^^-  9i/5. 

8.  Find  the  sum  of  2  V|,  V60,  VIB,  and  Vf. 


9.  Find  the  sum  of  A/45c3,  A/80c3,  and  A/5a*c. 

Ans.  (a-f  7c)A/5c. 
10.  Find  the  sum  of  Vl8a5^3  +  A/50a363. 

^ws.  (3a5 

^Vrf2 


11.  Find  the  sum  of 


/a2     ac      or/\      /7 

.     T  +  -T  +  —  JV7?- 
\5      d       m/  V  o 

12.  Find  the  sum  of  -/4a36,  -y/25^63,  and  5Z>  Vo&. 


RADICAL   QUANTITIES.  159 

To  find  the  Difference  of  Radical  Quantities, 
223.  When  the  radicals  are  similar,  it  is  evident  that  the 
subtraction  may  be  performed  in  the  same  manner  as  addition, 
except  that  the  signs  in  the  subtrahend  are  to  be  changed. 
Hence  we  have  the  following 

RULE. 

Reduce  each  radical  to  its  simplest  form.  If  the  resulting  radi 
cals  are  similar,  find  the  difference  of  the  coefficients,  and  to  the  re 
sult  annex  the  common  radical  part.  If  they  are  dissimilar,  flit 
subtraction  can  only  be  indicated. 

EXAMPLES. 

1.  From  V448  take  Vll2.  Ans.  4^7. 

2.  From  5  V20  take  3  V45.  Ans.  V5. 

3.  From  2\/50  take  Vl8.  Ans.  7 1/2. 

4.  From  VSOa^c  take  V20a*x3. 

5.  From  21/720*  take  Vl62a2. 

6.  From  v/192  take  v/24.  Ans.  2\/3. 

7.  From  21/320  take  3v/40. 


r>    -n  i  .         ,_         .. 

8.  From  X-^-  take  \-^.  J.TW.  (3a-l- 

2T?  multiply  Radical  Quantities  together. 

224.  We  have  found,  Art.  215,  that  Va  multiplied  by  Vb  is 
equal  to  Vab. 

Hence,  V%  X  VS  =  V$. 

If  the  radicals  have  coefficients,  the  product  of  the  coefE* 
cients  may  be  taken  separately. 

Thus,       aVxxl)Vy=axbxVxxVy=abVxy; 
also,  3-v/8x5V2^15Vl6  =  60. 

If  the  radicals  have  not  a  common  index,  they  must  first  be 
reduced  to  a  common  index.  Hence  we  have  the  following 


160  ALGEBRA. 

RULE. 

If  necessary,  reduce  the  given  radicals  to  a  common  index. 
tiply  the  coefficients  together  for  a  new  coefficient ;  also  multiply  the 
quantities  under  the  radical  signs  together,  and  place  this  product 
under  the  common  radical  sign.  Then  reduce  the  result  to  its  sim 
plest  form. 

EXAMPLES. 

Find  the  value  of  the  following  expressions : 

1.  3V8X2-/6.  Ans.  24\/3. 

2.  5V8X3V5.  Ans.  30VlO. 

3.  V2  x  \/3.  Ans.  \/72. 

4.  5  V3  x  7  VI  x  V2.  Ans.  140. 

5.  cVaxdVa.  Ans.  acd. 

6.  7v/18x5v/4.  Ans.  70v/9. 

7.  iV6xAl/l7. 

8.  i  1/18x61/20. 

9.  3y/4  x  7  V/6  x  J 1/6.  ^s.  7  v/15. 

225.  We  have  seen,  Art.  58,  that  the  exponent  of  any  letter  in 
a  product  is  equal  to  the  sum  of  the  exponents  of  this  letter  in  the 
multiplicand  and  multiplier.  That  is,  amxa"=am+n,  where  m 
and  n  are  supposed  to  be  positive  whole  numbers. 

When  one  or  both  of  the  exponents  are  negative,  we  must 
take  the  algebraic  sum  of  the  exponents.  For,  suppose  n  is 
negative.  Then 

am  x  a~n = am  x  -,  by  Art.  76,  =  ^ = am~n. 

The  same  relation  holds  true  when  m  and  n  are  fractional ; 

p         r         p+r 

that  is,  a*xas=a«  *. 

For  a*  x  as= \fa?  X  t/ar,  Art.  214,  ^V^  x  Va^,  Art.  220, 


Hence  we  conclude  that  the  exponent  of  any  letter  in  a  product 
is  equal  to  the  algebraic  sum  of  the  exponents  of  this  letter  in  the 


RADICAL  QUANTITIES.  161 

multiplicand  and  multiplier,  whether  the  exponents  are  positive  or 
negative,  integral  or  fractional. 

EXAMPLES. 

1.  Multiply  5a*  by  3a*.  Arts.  Ufa* 

2.  Multiply  2la*  by  3a8.  Ans.  63a^. 

3.  Multiply  8x*y*  by  4a:*y* 

4.  Find  the  product  of  a2,  a  ,  a  ,  and  cT^*. 

i     _.&     4  1 

5.  Find  the  product  of  a  ,  a     ,  a^,  and  a1"2". 

Multiplication  of  Polynomial  Radicals. 

226.  By  combining  the  preceding  rules  with  that  for  the 
multiplication  of  polynomials,  Art.  61,  we  may  multiply  togeth 
er  radical  expressions  consisting  of  any  number  of  terms. 

Ex.  1.  Let  it  be  required  to  multiply 


-3a  - 


Ex.  2.  Multiply  3  +  V5  by  2  —  VS.  Ans.  1  —  V5. 

Ex.  3.  Multiply  7  +  2  V6  by  9-5  V6.         Ans.  3- 17  VS. 
Ex.  4.  Multiply  9  +  2  VlO  by  9-2  VlO.  Ans.  41. 

Ex.  5.  Multiply  3 V45-7 V5  by  Vl|+2  V§|.      Ans.  34. 

Ex.  6.  Multiply  cVa  +  dVb  by  cVa—dVb. 

A.ns,  ac  —  ooL . 


Ex.  7.  Multiply  aT— a3  +  a¥— a2+a^— a+a¥—  1  by  a 

.Ans.  a4 — 1. 

4* 


162  ALGEBKA. 

To  divide  one  Radical  Quantity  by  another. 

227.  The  division  of  radical  quantities  depends  upon  the  fol 
lowing  principle  : 

The  quotient  of  the  nth  roots  of  two  quantities  is  equal  to  the  nth 
root  of  their  quotient  ;  or, 


for  the  nth  power  of  each  of  these  expressions  is  ^,  Art.  186. 

Let  it  be  required  to  divide  4a2A/6fo/  by  2aVSb. 

4a2    /%  /— 

—\/-^=2aV2y.  Ans. 


2aVSb 
Hence  we  have  the  following 

EULE. 

If  necessary,  reduce  the  given  radicals  to  a  common  index.  Di 
vide  the  coefficient  of  the  dividend  by  that  of  the  divisor  for  a  new 
coefficient  ;  also  the  quantity  under  the  radical  sign  in  the  dividend 
by  that  in  the  divisor,  and  place  this  quotient  under  the  common 
radical  sign.  Then  reduce  the  result  to  its  simplest  form. 

EXAMPLES. 

1.  Divide  8VT08  by  2V  6.  Ans.  12  1/2. 

2.  Divide  8  v/512  by  4  i/2.  Ans.  8  3v/4. 

3.  Divide  6  3/54  by  3  v/2.  Ans.  6. 

4.  Divide  4  v/72  by  2  v/18. 

5.  Divide  4\/6a2?/  by 


.  _ 

6.  Divide  16(a36)w  by  8(ac)m. 

7.  Divide  4  v/12  by  2  V^.  Ans.  2 

8.  Divide  v/64  by  2. 

9.  Divide  Va2bc  by  \ 


228.  We  have  seen,  J.r£.  72,  that  /7ie  exponent  of  any  letter  in 


KADICAL   QUANTITIES.  163 

a  quotient  is  equal  to  the  difference  between  the  exponents  of  this  let 
ter  in  the  divisor  and  dividend. 

The  same  relation  holds  true  whether  the  exponents  are  posi 
tive  or  negative,  integral  or  fractional  ,  that  is,  universally, 


For  the  quotient  must  be  a  quantity  which,  multiplied  by 
the  divisor,  shall  produce  the  dividend  ;  and,  according  to  Art. 
225,  the  exponent  of  any  letter  in  a  product  is  in  all  cases 
equal  to  the  algebraic  sum  of  the  exponents  of  this  letter  in 
the  multiplicand  and  multiplier.  Hence  this  relation  must 
hold  true  universally  in  division. 

EXAMPLES. 

1.  Divide  (ab)3  by  (o5)*  Ans.  (afy*. 

2,.  1 

2.  Divide  a¥  by  a*. 

3.  Divide  4:Vab  by  2  \/ab.  Ans.  2  Vab. 

4.  Divide  9m2(a—  1$  by  Sm(a—  bfi.        Ans.  Sm(a-b)^. 

1  1  JL  JL 

5.  Divide  a2&3  by  aBb5.  Ans.  a 


6.  Divide  4*  by  2*.  Ans,  4= 

V2 

Division  of  Polynomial  Radicals. 

229.  By  combining  the  preceding  rules  with  that  for  the  di 
vision  of  polynomials,  Art.  80,  we  may  divide  one  radical  ex 
pression  by  another  containing  any  number  of  terms. 

A        .2.         1         1  1 

Ex.  1.  Let  it  be  required  to  divide  a6  —  a3  —  a2  -far  by  a^— 1. 

$  I  ai_i 


Ex.2.  Divide  8a-6  by  2a~ 


164  ALGEBKA. 

Ex.  3.  Divide  a-41o^-120  by  a^ 

Ans.  a~z — 
Ex.  4.  Divide  a*+64&*  by  a^+46*. 

Ans.  a?— 4a^6^"-f  16&*. 

Ex.  5.  Divide  x%— xy^+x^y— y^  by  x^—y^. 

,  Ans.  x+y. 

To  involve  a  Radical  Quantity  to  any  power. 

^ 
230.  Let  it  be  required  to  raise  am  to  the  nth  power. 

Ill      l_i_l       JL 
The  square  of  am  is  amxam=am  m=am. 

1         til       1+1+1       A 
The  cube  of  am  is  amxamxam=a™^~m=an. 

1111  -+-+-  etc        H 

The  nth  power  of  am  is  am  x  am  x  am,  etc.,  =am  m  m'    '' =am. 

Hence,  to  involve  a  radical  quantity  to  any  power,  we  have 
the  following 

RULE. 

Multiply  the  fractional  exponent  of  the  quantity  by  the  exponent 
of  the  required  power.  If  the  radical  has  a  coefficient,  let  this  be  in 
volved  separately ;  then  reduce  the  result  to  its  simplest  form. 

If  the  quantity  is  under  the  radical  sign,  it  is  generally  most 
convenient  to  substitute  for  this  sign  the  equivalent  fractional 
exponent ;  but  if  we  choose  to  retain  the  radical  sign,  we  must 
raise  the  quantity  under  it  to  the  required  power. 

EXAMPLES. 

1  4 

1.  Eequired  the  fourth  power  of  fa*.  Ans. 

2.  Eequired  the  cube  of  |V3.  Ans.  f 

3.  Eequired  the  square  of  3  v/3. 

4.  Eequired  the  cube  of  17V21. 

5.  Eequired  the  fourth  power  of  iV6.  Ans. 

6.  Eequired  the  fourth  power  of  2  v/3o*&. 

j 

7.  Eequired  the  fourth  power  of  abVab.  Ans.  a6b*. 


RADICAL  QUANTITIES.  165 

8.  Kequired  the  sixth  power  of  (a+b) *. 

Ans.  a2  +  2ab  +  b\ 

9.  Eequired  the  value  of  V(l|)7  x  V(£f)6.  Ans.  -&. 

10.  Kequired  the  value  of  V(±ab2)x  x  V(2a?by. 

Ans.  (2ab)*. 

To  Extract  any  Root  of  a  Radical  Quantity. 
231.  A  root  of  a  quantity  is  a  factor  which,  multiplied  by 
itself  a  certain  number  of  times,  will  produce  the  given  quanti- 

-I        * 
ty.     But  we  have  seen  that  the  nth  power  of  am  is  am.     There- 

n^  _!_ 

fore  the  nth  root  of  am  is  am.     Hence  we  derive  the  following 

RULE. 

Divide  the  fractional  exponent  of  the  quantity  by  the  index  of  the 
required  root.  If  the  radical  has  a  coefficient,  extract  its  root  sep 
arately  if  possible ;  otherwise  introduce  it  under  the  radical  sign. 
Then  reduce  the  result  to  its  simplest  form. 

If  the  quantity  is  under  the  radical  sign,  and  we  choose  to 
retain  the  sign,  we  must,  if  possible,  extract  the  required  root 
of  the  quantityunder  the  radical  sign  ;  otherwise  we  must  mul 
tiply  the  index  of  the  radical  by  the  index  of  the  required  root. 

EXAMPLES. 

1.  Find  the  square  root  of  9(3)^.  Ans.  3  v/3. 

2.  Find  the  cube  root  of  $V5.  Ans.  i  6/2. 
8.  Find  the  square  root  of  103. 

4.  Find  the  cube  root  of 

5.  Find  the  fourth  root 

6.  Find  the  cube  root  of 

7.  Find  the  cube  root  #  | \/|  Ans-  \/f 

8.  Find  the  square  root  of  3  v/5.  Ans.  y/135. 

9.  Find  the  fourth  root  of  %  v/|.  Ans.  $  \/l2. 


166  ALGEBRA. 

Operations  on  Imaginary  Quantities. 

232.  It  has  been  shown,  Art.  195,  that  an  even  root  of  a  neg 
ative  quantity  is  impossible.     Thus,  V—  4,  V  —  9,  V—  5a  are 
algebraic  symbols  representing  operations  which  it  is  impossi 
ble  to  execute  ;  for  the  square  of  every  quantity,  whether  posi 
tive  or  negative,  is  necessarily  positive.     Quantities  of  this  na* 
ture  are  called  imaginary  or  impossible  quantities.     Neverthe 
less,  such  expressions  do  frequently  occur,  and  it  is  necessary 
to  establish  proper  rules  for  operating  upon  them. 

233.  The  square  root  of  a  negative  quantity  may  always  be  rep 
resented  by  the  square  root  of  a  positive  quantity  multiplied  by  the 
square  root  of  —I. 

Thus  1/^4  =  V±x  -1  = 


The  factor  V  —  1  is  called  the  imaginary  factor,  and  the  other 
factor  is  called  its  coefficient. 

234.  When  several  imaginary  factors  are  to  be  multiplied 
together,  it  is  best  to  resolve  each  of  them  into  two  factors,  of 
which  one  is  the  square  root  of  a  positive  quantity,  and  the 
other  V  —  1.  We  can  then  multiply  together  the  coefficients 
of  the  imaginary  factor  by  methods  already  explained.  It 
only  remains  to  deduce  a  rule  for  multiplying  the  imaginary 
factor  into  itself;  that  is,  for  raising  the  imaginary  factor  to  a 
power  whose  exponent  is  equal  to  the  number  of  factors. 

The  first  power  of  V—  1  is  V—  1. 

The  second  power,  by  the  definition  of  square  root,  is  —  1. 

The  third  power  is  the  product  of  the  first  and  second  pow 

ers,  or  —  Ix  V—  1  =  —  V  —  1. 

The  fourth  power  is  the  square  of  the  second,  or  +1. 

The  fifth  is  the  product  of  the  first  and  fourth  ;  that  is,  it  is 
the  same  as  the  first;  the  sixth  is  the  same  as  the  second,  and 


RADICAL   QUANTITIES. 


167 


so  on ;  so  that  all  the  powers  of  V— 1  form  a  repeating  cycle 
of  the  following  terms : 

EXAMPLES. 

1.  Multiply  V  —  9  by  V— 4. 

V^x  V^4  =  3V^Tx2V^I=£ 

2.  Multiply  1+V^T  by  1—  V^T. 

3.  Multiply  Vl8  by  V^2. 

4.  Multiply  5  +  2V^3  by  2-V^1 

5.  Multiply  aV  —  b  by  cV—d. 

6.  Multiply  1-V^l  by  itself. 

7.  Multiply  2V3—  V— 5  by  4 A/3- 

JL    «/  »/ 

J.W5.    1 

8.  Multiply  a+  VbV  —  1  by  a—  VbV^  1. 

9.  Multiply  aV  —  a2b3  by   V  — a4^5. 

10.  Multiply  V  —  a-\-V~^b  by  V— a— y^6. 


u4ws.  — acVbd. 
Ans.  —2V  —  1. 


235.  Division  of  Imaginary  Quantities. — The  quotient  of  one 
imaginary  term  divided  by  another  is  easily  found  by  resolv 
ing  both  terms  into  factors,  as  in  the  preceding  article. 

Ex.  1.  Let  it  be  required  to  divide  V^ab  by   V  —  a. 


—  =  y  6,  A7Z5. 


—  a 


Ex.  2.  Divide  Z>V^1  by  cV— 1, 

Ex.  3.  Divide  unity  by  V^T. 
Ex.  4.  Divide  a  by  6V— 1. 
Ex.  5.  Divide  a  by  V«V— 1. 


.  —  V  —  1. 


168  ALGEBRA. 


Ex.  6.  Divide  V^l2  +  V^6+V^9  by 
Ex.  7.  Divide  2V8-V-10  by  --/^2. 

Ans 


Multipliers  which  shall  cause  Surds  to  become  Rational. 

236.  1st.  When  the  surd  is  a  monomial. 

The  quantity  Va  is  rendered  rational  by  multiplying  it  by 

Va. 

i       i 

For  -y/a  X  Va—o^  x  a^—a. 

1  2 

So,  also,  aT  is  rendered  rational  by  multiplying  it  by  a7. 

i  -3  i 

Also,  a4  is  rendered  rational  by  multiplying  it  by  a4  ;  and  an 

by  multiplying  it  by  a     n. 

Hence  we  deduce  the  following 

RULE. 

Multiply  the  surd  l>y  the  same  quantity  having  such  an  exponent 
as,  when  added  to  the  exponent  of  the  given  surd,  shall  make  unity. 

237.  2d,  When  the  surd  is  a  binomial. 

If  the  binomial  contains  only  the  square  root,  multiply  the 
given  binomial  by  the  same  terms  connected  by  the  opposite  sign, 
and  it  will  give  a  rational  product. 

Thus  the  expression  Va-\-Vb  multiplied  by  Va—Vb  gives 
for  a  product  a—  b. 

Also  the  expression  \/a+Vb  multiplied  by  \/a—  Vb  gives 
for  a  product  Va—Vb,  which  may  be  rendered  rational  by 
multiplying  it  by  Va+Vb. 

In  general,  m\/a  ±  Vb  may  be  rendered  rational  by  success 
ive  multiplications  whenever  m  and  n  denote  any  power  of  2. 
When  m  and  n  are  not  powers  of  2,  the  binomial  may  still  be 
rendered  rational  by  multiplication,  but  the  process  becomes 
more  complicated. 

Ex.  1.  Find  a  multiplier  which  shall  render  1/6  +  VB  ration 
al,  and  determine  the  product. 


RADICAL   QUANTITIES.  169 

Ex.  2.  Find  a  multiplier  which  shall  render  A/3  —  Vx  ration 
al,  and  determine  the  product. 

Ex.  3.  Find  multipliers  which  shall  render  A/3  — v/^  ration 
al,  and  determine  the  product. 

» 

238.  3d.  When  the  surd  is  a  trinomial 

When  a  trinomial  surd  contains  only  radicals  of  the  second 
degree,  we  may  reduce  it  to  a  binomial  surd  by  multiplying  it 
by  the  same  expression,  with  the  sign  of  one  of  the  terms 
changed.  Thus,  Va+Vb+Vc  multiplied  by  Va+Vb—Vc 
gives  for  a  product  a  +  b  —  c  +  2Vabi  which  may  be  put  under 
the  form  of  m-{-2Vab. 

Ex.1.  Find  multipliers  that  shall  make  A/5 -f  A/3— A/2  ra 
tional,  and  determine  the  product. 

Ex.  2.  Find  multipliers  that  shall  make  1-f-  A/2  +  A/3  ration 
al,  and  determine  the  product. 

To  transform  a  Fraction  whose  Denominator  is  a  Surd  in  such 
a  Manner  that  the  Denominator  shall  be  Rational. 

239.  If  we  have  a  radical  expression  of  the  form 


a  Vb+Vc 

or  7/7     77=,  it  may  be  transformed  into  an  equivalent  expres 

sion  in  which  the  denominator  is  rational  by  multiplying  both 
terms  of  the  fraction  by  Vb±  Vc.     Hence  the 

RULE. 

Multiply  both  numerator  and  denominator  by  a  factor  which 
ivill  render  the  denominator  rational. 

EXAMPLES. 

Reduce  the  following  fractions  to  equivalent  fractions  having 
a  rational  denominator  : 

2  2A/3 

L 


2.  _  Ans. 

A/5-  A/2 

IT 


170  ALGEBKA. 


3.    --  -=.  Ans. 

3- 


4. 


a+Vb 
5. 


8.     ~1  Ans.')/*. 

1+a+Vl  —  a2  ,        l  +  Vl—a2 

1+a—Vl  —  a2'  & 

240.  The  utility  of  the  preceding  transformations  will  be  seen 
if  we  attempt  to  compute  the  numerical  value  of  a  fractional 
surd. 

Ex.  1.  Let  it  be  required  to  find  the  square  root  of  T ;  that  is, 

Vs 

to  find  the  value  of  the  fraction  — . 

Making  the  denominator  rational,  we  have      ^    ,  and  the 

value  of  the  fraction  is  found  to  be  0.6546. 

7V5 

Ex.  2.  Compute  the  value  of  the  fraction  — — . 

Vll+VS 

Ans.  3.1003. 

•v/6 
Ex.  3.  Compute  the  value  of  the  fraction  — — . 

V7-h  V3 

Ans.  0.5595. 

1/8 

Ex.  4.  Compute  the  value  of  the  fraction  — -=—  —j=. 

2v8+3V5-7v2 

Ans.  0.7025. 


RADICAL   QUANTITIES.  171 

94-2A/IO 


Ex.  5.  Compute  the  value  of  the  fraction 


9-2  VlO 

Ans.  5.7278. 


Square  Root  of  a  Binomial  Surd. 

241.  A  binomial  surd  is  a  binomial,  one  or  both  of  whose 
terms  are  surds,  as  2-f  V3  and  1/5—  A/2. 

A  quadratic  surd  is  the  square  root  of  an  imperfect  square. 

If  .we  square  the  binomial  surd  2+  A/3,  we  shall  obtain 
7  +  4A/3.  Hence  the  square  root  of  7+4V3  is  2+  A/3;  that 
is,  a  binomial  surd  of  the  form  a  ±  Vl>  may  sometimes  be  a  perfect 
square. 

242.  The  method  of  extracting  the  square  root  of  an  expres 
sion  of  the-  form  a±  Vb  is  founded  upon  the  following  princi 
ples: 

1st.  The  sum  or  difference  of  two  quadratic  surds  can  not  be 
equal  to  a  rational  quantity. 

Let  A/a  and  Vb  denote  two  surd  quantities,  and,  if  possible, 


where  c  denotes  a  rational  quantity.     By  transposing  Vb  and 
squaring  both  members,  we  obtain 


The  second  member  of  the  equation  contains  only  rational 
quantities,  while  Vb  was  supposed  to  be  irrational;  that  is,  we 
have  an  irrational  quantity  equal  to  a  rational  one,  which  is 
impossible.  Hence  the  sum  or  difference  of  two  quadratic 
surds  can  not  be  equal  to  a  rational  quantity. 

243.  2d.  In  any  equation  which  involves  both  rational  quanti 
ties  and  quadratic  surds,  the  rational  parts  in  the  two  members  are 
equal,  and  also  the  irrational  parts. 

Suppose  we  have 


172  ALGEBRA. 

Then,  if  x  be  not  equal  to  a,  suppose  it  to  be  equal  to  a-f  m; 

then  /-  /7 

y  —  a-\-  Vb, 


so  that  m=Vb—Vy; 

that  is,  a  rational  quantity  is  equal  to  the  difference  of  two 
quadratic  surds,  which,  by  the  last  article,  is  impossible.  There 
fore  x=a,  and  consequently  Vy—Vb. 

244.  To  find  an  expression  for  the  square  root  of  a±  Vb. 


Let  us  assume         Va  +  Vb  —  Vx  +  Vy.  (1.) 

By  squaring,         a+  Vb—x-\-  2Vxy+y.  (2.) 

By  Art.  243,                    a=x+y,  (3.) 

and                                     Vb  =  2Vxy.  (4.) 
Subtracting  (4)  from  (3),  we  have 

a—Vb=x-2Vxy+y.  (5.) 

By  evolution,         Va—Vb  =Vx—Vy.  (6.) 
Multiplying  (1)  by  (6),  we  have 

(7.) 

Adding  (3)  and  (7),  a+V~a^-b  =  2x.  (8.) 


TT-  a-hvcfc2— u  /n  \ 

Hence,  cc  = —  (9.) 

Subtracting  (7)  from  (3), 


It  is  obvious  from  these  equations  that  x  and  y  will  be  ra 
tional  when  az  —  b  is  a  perfect  square.  If  «2  —  b  be  not  a  perfect 
square,  the  values  of  ~Jx  and  Vy  will  be  complex  surds. 

Hence,  to  obtain  the  square  root  of  a  binomial  surd,  we  pro 
ceed  as  follows : 

Let  a  represent  the  rational  part,  and  Vb  the  radical  part, 
and  find  the  values  of  x  and  y  in  equations  (9)  and  (10).  Then, 
if  the  binomial  is  of  the  form  a-fVS,  its  square  root  will  be 
Vx+Vy.  If  the  binomial  is  of  the  form  a—  Vb,  its  square 
root  will  be  ^fx—  V y.  ^ 


RADICAL   QUANTITIES. 
EXAMPLES. 

1.  Required  the  square  root  of  4  +  2V3. 
Here  a =4  and  Vb  = 


Hence 


2 
Hence  Vx+  Vy=  V3  +  1,  Ans. 

Verification.  The  square  of  VB  +  1  is  3  +  2/84-1=4+21/3. 

2.  Kequired  the  square  root  of  11  +  6V2. 
Here  a=ll  and  Vft  =  6-v/2;   or  6  =  72. 

ce=9  and  y=2. 
Vx+Vy  =  3+V%,  Ans. 

3.  Required  the  square  root  of  11-2/30.     Ans.  -v/6—  V5. 

4.  Required  the  square  root  of  2+  Vs.         Ans.  V|+  V^» 

5.  Required  the  square  root  of  7  +  2  VTO.     Ans.  V5+V2. 

6.  Required  the  square  root  of  18  +  8V5. 

Ans.  VIO  +  2-V/2. 

245.  This  method  is  applicable  even  when  the  binomial  con 
tains  imaginary  quantities. 

7.  Required  the  square  root  of  1+4V—  3. 
Here  a  =  l  and  V£=4V^3. 

Hence  6=  -48  and  a2-&=49. 

Therefore  x=4:  and  y=—  3. 

The  required  square  root  is  therefore  2  +  V  —  3, 

8.  Required  the  square  root  of  —  2-+|V—  3. 


9.  Required  the  square  root  of  2V  —  1  or 

Ans. 


174  ALGEBRA. 

10.  Eequired  the  value  of  the  expression 

V6  +  2V5-V6-2V5. 
IL  Eequired  the  value  of  the  expression 

V4  +  3V^H-  V4-3V-20.  <4ws.  6. 

12.  Eequired  the  square  root  of  —  3+  V  —  16. 

Ans. 

13.  Eequired  the  square  root  of  8  A/  —  1, 


Simple  Equations  containing  Radical  Quantities. 
246.  When  the  unknown  quantity  is  affected  by  the  radical 
sign,  we  must  first  render  the  terms  containing  the  unknown 
quantity  rational.  This  may  generally  be  done  by  successive 
involutions.  For  this  purpose  we  first  free  the  equation  from 
fractions.  If  there  is  but  one  radical  expression,  we  bring  that 
to  stand  alone  on  one  side  of  the  equation,  and  involve  both 
members  to  a  power  denoted  by  the  index  of  the  radical. 


Ex.1.  Given  x-}-  Vx2-3x  +  60  =  12  to  find  x. 
Transposing  x  and  squaring  each  member,  we  have 


whence  x =4. 

2a 
.   Given  Vx+  va+x=— -p =  to  find  x. 

I/ /-»        I       SY> 

Ans.  x=%. 


Ex.3.  Given 

Ex,  4.  Given  V4x2— Ix—  6  =  9  —  2x  to  find  x. 

247.  If  the  equation  contains  two  radical  expressions  com 
bined  with  other  terms  which  are  rational,  it  is  generally  best 
to  bring  one  of  the  radicals  to  stand  alone  on  one  side  of  the 
equation  before  involution.  One  of  the  radicals  will  thus  be 
made  to  disappear,  and,  by  repeating  the  operation,  the  remain 
ing  radical  may  be  exterminated. 

Ex.5.  Given  Vx-}-19+  Vx+lO  —  Q  to  find  x. 


RADICAL   QUANTITIES. 


175 


By  transposition, 

Squaring, 

Transposing  and  reducing, 

VoH-10=4. 

Squaring,  x+ 10  =  16 ; 

whence  x  — 6. 

Ex.6.  Given  V36+x=18-h Vx  to  find  x. 

Ex.7.   Given  Vx  +  ±ab  =  2b  +  Vx  to  find  x. 

Ex.  8.  Given  x—  Va2+xVb*+x2  —  a?+a  to  find  x. 

248.  When  an  equation  contains  a  fraction  involving  radical 
quantities  in  both  numerator  and  denominator,  it  is  sometimes 
best  to  render  the  denominator  rational  by  Art.  239 ;  but  the 
best  method  can  only  be  determined  by  trial. 


Ex.9.  Given 


to  find  x. 


Multiply  both  terms  of  the  first  fraction  by  Vx+  Vx—3, 
and  we  have  (Vx+  Vx—  3)2       3 


or 


:x-3' 


Extracting  the  square  root, 

y^ 
Clearing  of  fractions, 


whence 

Ex.  10.  Given 

Ex.  11.  Given 

Ex.  12.  Given 


176  ALGEBKA. 

Ex.  13.  Given     S^~l  =l  +  £(V&c-l)  to  find  x. 


Ex.  14.  Given  -^  *  to  find  x. 

,.  a?=/r  my*  V 

\m  —  n) 

Ex.15.  Given  (VSc-6)  (Vie  +  25)=(5  +  3VaO  (Vx+3)  to 
find  x.  Ans.  x  =  9. 

Ex.16.  Given  V'2x— 3n  =  3V?i—  V%x  to  find  x. 

*iLns.  x  —  An, 


^     17    p.  - 

Ex.17.  Given  --=  -  —     ._        -  to  find  x. 
Vx+2 


—  - 

Ex.18.  Given  —  =  -  =  —  —     -  to  find  x. 


Ex.19.  Given  --l^Z      to  find  x. 


Ex.  20.  Given   /4a+x=2  Vb+x—  Vx  to  find  x. 

(a- 
Ans.  x=: 


:  -- 
2a—  6 


EQUATIONS  OF  THE   SECOND   DEGREE.  177 


CHAPTER  XIV. 

EQUATIONS  OF  THE   SECOND   DEGREE. 

249.  An  equation  of  the  second  degree,  or  a  quadratic  equation 
with  one  unknown  quantity,  is  one  in  which  the  highest  power 
of  the  unknown  quantity  is  a  square. 

250.  Every  equation  of  the  second  degree  containing  but 
one  unknown  quantity  can  be  reduced  to  three  terms;  one  con 
taining  the  second  power  of  the  unknown  quantity,  another  the 
first  power,  and  the  third  a  known  quantity  ;  that  is,  it  can  be 
reduced  to  the  form  _ 


Suppose  we  have  the  equation 


Clearing  of  fractions  and  expanding,  we  have 

9x2+7x-6  +  4x-8  +  164:=12x2+24x-36. 
Transposing  and  uniting  similar  terms,  we  have 

-3x2-13x=-186. 
Dividing  by  the  coefficient  of  x2,  that  is,  by  —3,  we  have 


which  is  of  the  form  above  given,  p  in  this  case  being  equal  to 
^,  and  q  being  equal  to  62. 

251.  In  order  to  reduce  a  quadratic  equation  to  three  terms,  we 
must  first  clear  it  of  fractions,  and  perform  all  the  operations 
indicated.  We  then  transpose  all  the  terms  which  contain  the 
unknown  quantity  to  the  first  member  of  the  equation,  and  the 
known  quantities  to  the  second  member;  unite  similar  terms, 
and  divide  each  term  of  the  resulting  equation  by  the  coeffi 
cient  of  x2. 


178  ALGEBRA. 

252.  An  equation  of  the  form  x2+px=q  is  called  a  complete 
equation  of  the  second  degree,  because  it  contains  each  class  of 
terms  of  which  the  general  equation  is  susceptible. 

253.  The  coefficient  of  the  first  power  of  the  unknown  quan 
tity  may  reduce  to  zero,  in  which  case  the  equation  is  said  to 
be  incomplete. 

An  incomplete  equation  of  the  second  degree,  when  reduced, 
contains  but  two  terms :  one  containing  the  square  of  the  un 
known  quantity,  and  the  other  a  known  term. 

Incomplete  Equations  of  the  Second  Degree. 

254.  An  incomplete  equation  of  the  second  degree  may  be 
reduced  to  the  form  xz=q. 

Extracting  the  square  root  of  each  member,  we  have 
x=±i/q. 

If  q  be  a  positive  number,  either  integral  or  fractional,  we 
can  extract  its  square  root,  either  exactly  or  approximately,  by 
the  rules  of  Chap.  XII.  Hence,  to  solve  an  incomplete  equa 
tion  of  the  second  degree,  we  have  the 

RULE. 

Reduce  the  equation  to  the  form  xz=q,  and  extract  the  square 
root  of  each  member  of  the  equation. 

255.  Since  the  square  of  both   -fm  and  —  m  is   -\-m2,  the 
square  of  -f  Vq  and  that  of  —  Vq  are  both  +  q.     Hence  the 
above  equation  is  susceptible  of  two  solutions,  or  has  two  roots; 
that  is,  there  are  two  quantities,  which,  when  substituted  for  x 
in  the  original  equation,  will  render  the  two  members  identical. 
These  are  +Vq  and  —  Vq. 

Hence,  Every  incomplete  equation  of  the  second  degree  has  two 
roots,  equal  in  numerical  value,  but  with  opposite  signs. 

EXAMPLES. 

Find  the  values  of  x  in  each  of  the  following  equations: 
1.  4x2— 7  =  3z2-}-9.  Ans.  x=±4. 


EQUATIONS  OF  THE   SECOND   DEGREE.  179 

Show  that  each  of  these  values  will  satisfy  the  equation. 

Ans.  x=±7. 


±Vab 
Ans.  x=  —  ~  —  • 


__       2 
*        ~     f"~      f 


10.  xa4-Vx2-17=- 


Ans.x=±S. 


jZVbfe.  Clearing  of  fractions  and  transposing,  we  find  in  each 
member  of  this  equation  a  binomial  factor,  which  being  cancel 
ed,  the  equation  is  easily  solved. 

¥ 

PROBLEMS. 

Prob.  1.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  10  to  7,  and  whose  sum,  multiplied  by  the  less,  pro 
duces  270? 
,    Let  10x=  their  sum. 

Then  7x  —  the  greater  number, 

and  3x  =  the  less. 

Whence  30x2  =  270, 

and  x2  =  9; 

therefore  a;  =±3, 

and  the  numbers  are  ±21  and  ±9. 


180  ALGEBRA. 

Prob.  2.  What  two  numbers  are  those  whose  sum  is  to  the 
greater  as  m  to  n,  and  whose  sum,  multiplied  by  the  less,  is 
equal  to  a  ? 

A  /     an2  /a(m  —  n 

Ans.  ±  V  —/ x  and 

v  m(m  —  n) 

Prob.  3.  What  number  is  that,  the  third  part  of  whose 
square  being  subtracted  from  20,  leaves  a  remainder  equal 
to  8? 

Prob.  4.  What  number  is  that,  the  rath  part  of  whose  square 
being  subtracted  from  a,  leaves  a  remainder  equal  to  b  ? 

Ans.  ±  Vm(a  —  b). 

12          3 
Prob.  5.  Find  three  numbers  in  the  ratio  of  ^,  ^  and  7,  the 

£     O  4: 

sum  of  whose  squares  is  724. 

Prob.  6.  Find  three  numbers  in  the  ratio  of  m,  n,  and  jo,  the 
sum  of  whose  squares  is  equal  to  a. 
Ans. 


am?  an2 


Prob.  7.  Divide  the  number  49  into  two  such  parts  that  the 
quotient  of  the  greater  divided  by  the  less  may  be  to  the  quo 
tient  of  the  less  divided  by  the  greater  as  |  to  -f . 

Ans.  21  and  28. 

Note.  In  solving  this  Problem,  it  is  necessary  to  assume  a 
principle  employed  in  Arithmetic,  viz.,  If  four  quantities  are 
proportional,  the  product  of  the  extremes  is  equal  to  the  product  of 
the  means. 

Thus,  if  a  :  b  ; :  c  :  d, 

then  ad=bc. 

Prob.  8.  Divide  the  number  a  into  two  such  parts  that  the 
quotient  of  the  greater  divided  by  the  less  may  be  to  the  quo 
tient  of  the  less  divided  by  the  greater  as  m  to  n.  , 

aVm           ,        aVn 
Ans.  — — and  — — -. 


m  -j-  Vn 
Prob.  9.  There  are  two  square  grass-plats,  a  side  of  one  of 


EQUATIONS  OF  THE  SECOND  DEGREE.         181 

which  is  10  yards  longer  than  a  side  of  the  other,  and  their 
areas  are  as  25  to  9.     What  are  the  lengths  of  the  sides? 

Prob.  10.  There  are  two  squares  whose  areas  are  as  m  to  T?, 
and  a  side  of  one  exceeds  a  side  of  the  other  by  a.  What  are 
the  lengths  of  the  sides  ? 

aVm  ,       aVn 

Ans.  —7= =  and  — 7=^ 


Vra  —  Vn  Vm  —  Vn 

Prob.  11.  Two  travelers,  A  and  B,  set  out  to  meet  each  other, 
A  leaving  Hartford  at  the  same  time  that  B  left  New  York. 
On  meeting,  it  appeared  that  A  had  traveled  18  miles  more 
than  B,  and  that  A  could  have  gone  B's  journey  in  15-|  hours, 
but  B  would  have  been  28  hours  in  performing  A's  journey. 
What  was  the  distance  between  Hartford  and  New  York? 

Ans.  126  miles. 

Projp.  12.  From  two  places  at  an  unknown  distance,  two 
bodies,  A  and  B,  move  toward  each  other,  A  going  a  miles  more 
than  B.  A  would  have  described  B's  distance  in  n  hours,  and 
B  would  have  described  A's  distance  in  m  hours.  What  was 
the  distance  of  the  two  places  from  each  other  ? 

Ans.  ax-—= 


— 

Prob.  13.  A  vintner  draws  a  certain  quantity  of  wine  out  of 
a  full  vessel  that  holds  256  gallons,  and  then,  filling  the  vessel 
with  water,  draws  off  the  same  quantity  of  liquor  as  before, 
and  so  on  for  four  draughts,  when  there  were  only  81  gallons 
of  pure  wine  left.  How  much  wine  did  he  draw  each  time? 

Ans.  64,  48,  36,  and  27  gallons. 

Note.  Suppose  -  part  is  drawn  each  time. 

256     256(x  _  1) 
Then  256  —  -  —  remains  after  the  first  draught. 

x  x 

256(V  _  I")2 
Similarly,  -      '  2       -  remains  after  the  second  draught,  and 

so  on. 

256  (x-  1)4 
Hence  ~r  —  -==81. 


182  ALGEBRA. 

Prob.  14.  A  number  a  is  diminished  by  the  nth  part  of  it 
self,  this  remainder  is  diminished  by  the  ?ith  part  of  itself,  and 
so  on  to  the  fourth  remainder,  which  is  equal  to  b.  Eequired 
the  value  of  n. 

A  Va 

Ans.  —~  -  -. 


Prob.  15.  Two  workmen,  A  and  B,  were  engaged  to  work 
for  a  certain  number  of  days  at  different  rates.  At  the  end  of 
the  time,  A,  who  had  played  4  of  those  days,  received  75  shil 
lings,  but  B,  who  had  played  7  of  those  days,  received  only 
48  shillings.  Now  had  B  only  played  4  days  and  A  played  7 
days,  they  would  have  received  the  same  sum.  For  how 
many  days  were  they  engaged?  Ans.  19  days. 

Prob.  16.  A  person  employed  two  laborers,  allowing  them 
different  wages.  At  the  end  of  a  certain  number  of  days,  the 
first,  who  had  played  a  days,  received  m  shillings,  and  the 
second,  who  had  played  b  days,  received  n  shillings.  Now  if 
the  second  had  played  a  days,  and  the  other  b  days,  they  would 
both  have  received  the  same  sum.  For  how  many  days  were 
they  engaged? 

bVm—aVn  , 
Ans.  —  —  --  j=r-  clays. 
Vm  —  Vn 

Complete  Equations  of  the  Second  Degree. 

256.  In  order  to  solve  a  complete  equation  of  the  second  de 
gree,  let  the  equation  be  reduced  to  the  form 


If  we  can  by  any  transformation  render  the  first  member  of 
this  equation  the  perfect  square  of  a  binomial,  we  can  reduce 
the  equation  to  one  of  the  first  degree  by  extracting  its  square 
root. 

Now  we  know  that  the  square  of  a  binomial,  x+a,  or  rr2-f 
2ax-\-a2,  is  composed  of  the  square  of  the  first  term,  plus  twice 
the  product  of  the  first  term  by  the  second,  plus  the  square  of 
the  second  term. 

Hence,  considering  x*+px  as  the  first  two  terms  of  the 


EQUATIONS  OF  THE  SECOND   DEGREE.  183 

square  of  a  binomial,  and  consequently  px  as  being  twice  the 
product  of  the  first  term  of  the  binomial  by  the  second,  it  is 

evident  that  the  second  term  of  this  binomial  must  be  ~. 


257.  In  order,  therefore,  that  the  expression  x2+px  may  be 
rendered  a  perfect  square,  we  must  add  to  it  the  square  of  this 

f) 

second  term  ^  ;  and  in  order  that  the  equality  of  the  two  mem 
bers  may  not  be  destroyed,  we  must  add  the  same  quantity  to 
the  second  member  of  the  equation.  We  shall  then  have 

p2  p* 

x2+px  +  ^-=q+^-. 

Taking  the  square  root  of  each  member,  we  have 


whence,  by  transposition,  x=  —  ^-±  y  q-\-  ^-. 

Thus  the  equation  has  two  roots:  one  corresponding  to  the 
plus  sign  of  the  radical,  and  the  other  to  the  minus  sign.  These 
two  roots  are 


258.  Hence,  for  solving  a  complete  equation  of  the  second 
degree,  we  have  the  following 

RULE. 

1st.  Reduce  the  given  equation  to  the  form  ofx'z-{-px=q. 

2d.  Add  to  each  member  of  the  equation  the  square  of  half  the  co 
efficient  of  the  first  power  of  x. 

3d.  Extract  the  square  root  of  loth  members,  and  the  equation 
will  be  reduced  to  one  of  the  first  degree,  which  may  he  solved  in  the 
usual  manner. 

259.  When  the  equation  has  been  reduced  to  the  form  x2  + 
pz=q,  its  two  roots  will  be  equal  to  half  the  coefficient  of  the  sec- 


184  ALGEBRA. 

ond  term,  taken  with  a  contrary  sign,  plus  or  minus  the  square 
root  of  the  second  member,  increased  by  the  square  of  half  the  coeffi 
cient  of  the  second  term. 

Ex.  1.  Let  it  be  required  to  solve  the  equation 
x2-10x=-16. 

Completing  the  square  by  adding  to  each  member  the  square 
of  half  the  coefficient  of  the  second  term,  we  have 


Extracting  the  root,  x—  5~  ±3. 

Whence  x=5±3=     8  or  2,  Ans. 

To  verify  these  values  of  x,  substitute  them  in  the  original 
equation,  and  we  shall  have 

82-10x8  =  64-80=:-16. 
Also,  22-10x2  =  4-20=-16. 

Ex.  2.  Solve  the  equation  2x2  +  8x—  20  =  70. 

Ans.  x=+5  or  —9. 
Ex.  3.  Solve  the  equation  3o;2— 


Eed  u  ci  n  g,  x2  —  x  =  —  f  . 

Completing  the  square,  x2—  x+^=^—  |=^V- 
Hence  #=J±i  =  -for-J,  Ans. 

Second  Method  of  completing  the  Square. 

260.  The  preceding  method  of  completing  the  square  is  al 
ways  applicable  ;  nevertheless,  it  sometimes  gives  rise  to  incon 
venient  fractions.  In  such  cases  the  following  method  may  be 
preferred.  Let  the  equation  be  reduced  to  the  form 

ax2  -{-bx  =  c, 

in  which  a  and  b  are  whole  numbers,  and  prime  to  each  other, 
but  c  may  be  either  entire  or  fractional. 

Multiply  each  member  of  this  equation  by  4a,  and  it  becomes 


Adding  b2  to  each  member,  we  have 

4a2x2  4-  4abx  +  b*  =  4ac  -f  b2, 

where  the  first  member  is  a  complete  square,  and  its  terms  are 
entire. 

Extracting  the  square  root,  we  have 
2ax+  b= 


EQUATIONS  OF  THE  SECOND   DEGREE.  185 

Transposing  b,  and  dividing  by  2a, 


which  is  the  same  result  as  would  be  obtained  by  the  former 
rule  ;  but  by  this  method  we  have  avoided  the  introduction  of 
fractions  in  completing  the  square. 

If  b  is  an  even  number,  -  will  be  an  entire  number  ;  and  it 

& 

would  have  been  sufficient  to  multiply  each  member  by  a,  and 

£2 

add  —  to  each  member.     Hence  we  have  the  following 

RULE. 

1st.  Reduce  the  equation  to  the  form  ax2  +  bx=c,  where  a  and  b 
are  prime  to  each  other. 

Zd.^Ifb  is  an  odd  number,  multiply  the  equation  by  four  times 
the  coefficient  ofx2,  and  add  to  each  member  the  square  of  the  coeffi 
cient  ofx. 

3d.  If  b  is  an  even  number,  multiply  the  equation  by  the  coeffi 
cient  ofx2,  and  add  to  each  member  the  square  of  half  the  coefficient 
ofx. 

Ex.  4.  Solve  the  equation  6x2—  ISx-—  6. 

Multiplying  by  4x6,  and  adding  IS2  to  each  member,  we 
have  144:c2-312x-f  169  =  169-144=25. 

Extracting  the  root,    I2x  —  13  =  ±  5. 
Whence  l2x=lS  or  8, 

and  x=%  or  J-. 

Ex.5.  Solve  the  equation  llOx2—  21a=  —  1. 

Multiplying  by  440,  and  adding  212  to  each  member,  we  have 

48400x2-  9240x+  441  =  1. 
Extracting  the  root,      220x-  21  =  ±  1. 
Whence  x=^  or  -f^. 

Ex.6.  Solve  the  equation  7£2-3£=160. 

Ans.  x=5  or  —  £*.. 

261.  Modification  of  the  preceding  Method.  —  The  preceding 
method  sometimes  gives  rise  to  numbers  which  are  unneces- 


186  ALGEBRA. 

sarily  large.  When  the  equation  has  been  reduced  to  the  form 
ax2+bx=c,  it  is  sufficient  to  multiply  it  by  any  number  which 
will  render  the  first  term  a  perfect  square.  Let  the  resulting 
equation  be 

m2x2-{-nx=q. 
The  first  member  will  become  a  complete  square  by  the  ad- 

(n  \2 
- —  1  ,  and  the  equation  will  then  be 
2m/ 


4m2     *     4m2' 
This  method  is  expressed  in  the  following 

RULE. 

Having  reduced  the  equation  to  the  form  ax2-\-bx=c,  multiply 
the  equation  by  any  quantity  (the  least  possible)  which  will  render 
the  first  term  a  perfect  square.  Divide  the  coefficient  of  x  in  this 
new  equation  by  twice  the  square  root  of  the  coefficient  of  x2,  and 
add  the  square  of  this  result  to  both  members. 
Ex.7.  Solve  the  equation  8x24-9x=99. 

3249 


16 
9  ,  57 


33 

x=S  or  —-5-,  Ans. 
o 

Ex.  8.  Solve  the  equation  16x2— 15^=34. 

15\2     2401 


8/        64 

17 
te=8  or  -T, 

x=2  or  —  r— ,  Ans. 
ID 


Ex.  9.  Solve  the  equation  12x2 

Solve  the  following  equations : 

Ex.10,    frc2— &o5  +  20J=42t.  Ans.  x  =  7  or  -6J. 

Ex.  11.    x2— £-40  =  170.  Ans.  x=15  or  -14. 

Ex.12.    3x2-h2x-9  =  76. 


EQUATION'S   OF  THE   SECOND  DEGREE.  187 

Ex.13.     x2 


,-,  0  -  -         0 

Ex.  14.    &-—  —--—=2. 

This  equation  reduces  to  x2—  15  %x  =—46. 

Ans.  x—  11  J  or  4. 

Ex.  15.    ^-1^=^=2.  Ans.  x=3  or  6. 

x         2x2 


x     x+L     x+2 
Ex.17.   x2-xV3=x-%VS. 


V3  +  3       1/3-1 

Ans.  x=  —  —  or  —  —  • 


„         '    x—  1     cc—3        2 

Ex.18.    -     -z  --  -——  ~.  Ans.  x=5  or  1. 

a:—  2     x—  4         3 

,-,-   _,„        x        a-\-x     5 

Ex.  19.    ---  —  —  =-.  ^n».  cc=a  or  —  2«, 

x       2 


Ex.  20.    x2—  (a  +  l)x  +  ab  =  Q.  Ans.  x  =  a  or  b. 

Ex.21.    (3x-25)(7cc4-29)  =  0.  Ans.  a=8£  or  -4f 

3x_2     2x-5     10  13       1 


Ex.  23.     x-lx- 


Ans.  &=8  or    . 


™    01     170     170       51  ,12 

Ex.24.  --  = pr.  Ans.  x-^  or  —  If. 

x      x+l     x+2 

-r,             a2 -4- ax -\- x2     a2  —  ax4-x2  ab 

Ex.  25.    


a+x  a  —  x          3a— 

c=—  3a  or  3a 


Equations  which  may  be  solved  like  Quadratics. 

262.  There  are  many  equations  of  a  higher  degree  than  the 
second,  which  may  be  solved  by  methods  similar  to  those  em 
ployed  for  quadratics.  To  this  class  belong  all  equations  which 
contain  only  two  powers  of  the  unknown  quantity,  and  in  which 


188  ALGEBRA. 

ike  greater  exponent  is  double  the  less.    Such  equations  are  of  the 
form  x*n+px*  =  q, 

where  n  may  be  either  integral  or  fractional. 

For  if  we  assume  y=xn,  then  2/2=ic2n,  and  this  equation 
becomes  y*+py=<i; 

whence 

Extracting  the  nth  root  of  each  member,  we  have 


Ex.1.  Solve  the  equation  x*—  I3x2=—  36. 
Assuming  x2=y,  the  above  becomes 


whence  2/=9  or  4. 

But,  since  xz=y,  x—±Vy. 

Therefore  x=  ±  V$  or  ±  VI. 

Thus  x  has  four  values,  viz.,  +3,  -3,  +  2,  -2. 
To  verify  these  values: 

1st  value,  (  +  3)4-13(+3)*=-36,  i.e.,  81-117^-36, 
2d  value,  (-3)4-  13(-3)2=-36,  i.e.,  81-117  =  -36. 
3d  value,  (+2)4-13(  +  2)2=  _36,  i.e.,  16-  52=  -36. 
4th  value,  (_2)4-13(-2)2=  -36,  i.e.,  16-  52=  -36. 

Ex.  2.  Solve  the  equation  x6-35x3=—  216. 
Assuming  x3=y,  the  above  becomes 


whence  y—^  or  8. 

Hence  x=Vy=3  or  2. 

This  equation  has  four  other  roots  which  can  not  be  de 
termined  by  this  process. 

Ex.  3.  Solve  the  equation  cc-f  4A/#  —  21. 

Assuming  Vx=y,  we  have 


EQUATIONS  OF  THE  SECOND  DEGKEE.        189 

whence  y=3  or  —  7. 

Therefore  a  =9  or  49. 

Although  the  square  toot  of  9  is  generally  ambiguous,  and 
may  be  either  +3  or  —8,  still,  in  verifying  the  preceding  val 
ues,  T/X  can  not  be  taken  equal  to  —3,  because  9  was  obtained 
by  multiplying  +3  by  itself.  For  a  like  reason,  ^/x  can  not 
be  taken  equal  to  +  7.  A  similar  remark  is  applicable  to  sev 
eral  of  the  following  examples. 

263.  The  same  method  of  solution  may  often  be  extended 
to  equations  in  which  any  algebraic  expression  occurs  with  two 
exponents,  one  of  which  is  double  the  other. 

Ex.4.   Solve  the  equation  (x2+  x)z-26(x2+x)=-I20. 
Assuming  x*-\-x=y,  this  equation  becomes 

2/2-26y=-120; 

whence  2/=20  or  6. 

We  have  now  the  two  equations, 


the  first  of  which  gives  x=  —  5  or  +4, 
and  the  second  gives      x=—  3  or  +  2. 

Thus  the  equation  has  four  roots, 

-5,  +4,  -3,  +2, 

and  any  one  of  these  four  values  will  satisfy  the  given  equa 
tion. 

Ex.5.  Solve  the  equation  Va+12+  tAc+12  =  6. 

We  find  z-f-  12  =  16  or  81. 

Hence  x=4  or  69. 

1    Ex.6.  Solve  the  equation  2cc2+  V2a;2+l  =  ll. 

This  equation  may  be  written 


Hence  2x2  +  l  =  9  or  16; 

therefore  x  =+2,  -2,  +  V7|-,  - 


190  ALGEBRA. 

264.  Equations  of  the  Fourth  Degree.  —  An  equation  of  the 
fourth  degree  may  often  be  reduced  to  an  equation  contain 
ing  the  first  and  second  powers  of  some  compound  quantity,  " 
with  known  coefficients,  in  the  following  manner*  Transpose 
all  the  terms  to  the  first  member;  then  extract  the  square 
root  to  two  terms,  and  see  if  the  remainder  (with  or  without^ 
the  absolute  term)  is  a  multiple  of  the  root  already  obtained.: 

Ex.7.   Solve  the  equation  x4-12x3  +  44cc2-48x 

We  may  proceed  as  follows: 

x4  -  12z3  +  44x2  -  48x  -  9009  =  0  (  x2  -  fa 


_ 
-  Qx  )     -I2x3  4-  44x2 


8x2-48z-9009, 
or  8(cc2-6x)-9009. 

Hence  the  given  equation  may  be  expressed  as  follows: 


Ans.  a?  =13  or  —  7,  or  3±3V—  10. 

Ex.  8.  Solve  the  equation  x4-2x3+x=132. 

Ans.  x=4:  or  —3,  or  J±|V—  43. 
Ex.  9.  Solve  the  equation  x4-f  4x2  =  12. 

Ans..  x=±V%  or  ±  V  —  6. 


Ex.  10.  Solve  the  equation  x6—  8^3= 

Ans.  x=3  or  -v/19. 

6.  3. 

Ex.11.   Solve  the  equation  x5+x5-756. 

Ans.  cc=243  or  —  v/28*. 
Ex.12.  Solve  the  equation  -Jcc6—  ^x3~  —  •£$. 

Ans.  a?= 


Ex.13.  Solve  the  equation  2xTr+3orT=2. 

Ans.  x=^  or  —8. 

Ex.14.   Solve  the  equation  %x—$Vx =  22£.  ^ 

Ans.  x =49  or  — — . 

y 
Ex.15.  Solve  the  equation  vTO+^-v/10+^=2. 

Ans.  x  =  6  or  —9. 


EQUATIONS  OF  THE  SECOND  DEGKEE.        191 

Ex.  16.  Solve  the  equation  x6H-20x3-10=59. 

Ans.  x=V&  or  V~^&. 
Ex.  17.  Solve  the  equation  3x2n-2x«4-  3  =  11. 

,/  Ans.  x=V2  or  V—  £• 

Ex.18.  Solve  the  equation  Vl+a;-x2-2(l-ha;-cc2)  =  -J. 


or 


Ex.  19.  Solve  the  equation  Va5-hVi=20. 

4.7ZS.  z=256  or  625. 


Ex.20.   Solve  the  equation  x4— 

Ans.  x=B  or  —  1,  or  l±-/^5. 
Ex.  21.   Solve  the  equation  cc2+5x+4=5v^2+5x+28. 

4ws.  x=4:  or  —9,  or  —  f±-i  V-51. 
Ex.  22.  Solve  the  equation  x2  +  5=2Vx2—  2 


Ans.  x—  1. 

Ex.23.   Solve  the  equation  (x+  Vx)*-(x+  Vx)2  =  20592. 

Ans.  x  —  ^  or  Ifr 
Ex.  24.   Solve  the  equation  x+  V25+cc=157. 

Ans.  ccml44  or  171. 
Ex.25.  Solve  the  equation  Vx—l=x—l. 

Ans.  x=I  or  2. 

265.  We  have  seen  that  every  equation  of  the  second  de 
gree  has  two  roots  ;  that  is,  there  are  two  quantities  which  , 
when  substituted  for  x  in  the  original  equation,  will  render 
the  two  members  identical.  In  like  manner,  we  shall  find 
that  every  equation  of  the  third  degree  has  three  roots,  an  equa- 
tion  of  the  fourth  degree  has  four  roots,  and,  in  general,  an 
equation  of  the  rath  degree  has  m  roots. 

Before  determining  the  degree  of  an  equation,  it  should  be 
freed  from  fractions,  from  negative  exponents,  and  from  the 
radical  signs  which  affect  its  unknown  quantities.  Several  of 
the  preceding  examples  are  thus  found  to  furnish  equations 


192  ALGEBRA. 

of  the  fourth  degree,  while  others  furnish  equations  of  the 
second  degree. 

The  above  method  of  solving  the  equation  x2n+pxn-=q  will 
not  always  give  us  all  of  the  roots,  and  we  must  have  recourse 
to  different  processes  to  discover  the  remaining  roots.  The 
subject  will  be  more  fully  treated  in  Chapter  XXL 

Problems  producing  Equations  of  the  Second  Degree. 

Prob.  1.  It  is  required  to  find  two  numbers  such  that  their 
difference  shall  be  8  and  their  product  240. 
Let  x=the  least  number. 
Then  will  0:4-8  =  the  greater. 
And  by  the  question,  x(cc+8)=x'2-\-8x= 240. 
Therefore  x =12,  the  less  number, 

x +8  =  20,  the  greater. 
Proof.    20  — 12  =  8,  the  first  condition. 

20x12  =  240,  the  second  condition. 

Prob.  2.  The  Keceiving  Eeservoir  at  Yorkville  is  a  rectan 
gle,  60  rods  longer  than  it  is  broad,  and  its  area  is  5500  square 
rods.  Eequired  its  length  and  breadth. 

Prob.  3.  What  two  numbers  are  those  whose  difference  is 

2a,  and  product  If  

Ans.  a±Va2  +  bj  and   —  a±  yet2  4-6. 

Prob.  4.  It  is  required  to  divide  the  number  60  into  two 
such  parts  that  their  product  shall  be  864. 

Let  x  =  one  of  the  parts. 

Then  will  60  -  x = the  other  part. 

And  by  the  question,  #(60— a?)  =  60#-o?2=864. 

The  parts  are  36  and  24,  Ans. 

Prob.  5.  In  a  parcel  which  contains  52  coins  of  silver  and 
copper,  each  silver  coin  is  worth  as  many  cents  as  there  are 
copper  coins,  and  each  copper  coin  is  worth  as  many  cents  as 
there  are  silver  coins,  and  the  whole  are  worth  two  dollars. 
How  many  are  there  of  each  ? 

Prob.  6.  What  two  numbers  are  those  whose  sum  is  2a  and 
Vroduct  b?  Ans.  a+Va2  —  b  and  a-Va*-b. 


EQUATIONS   OF  THE   SECOND   DEGEEE.  193 

Prob.  7.  There  is  a  number  consisting  of  two  digits  whose 
sum  is  10,  and  the  sum  of  their  squares  is  58.  Kequired  the 
number. 

Let  x=the  first  digit. 

Then  will  10—  x=the  second  digit. 

And  x2+(10-*)2  =  2*2-2 

that  is  x2  —  I0x=  —  21 


=  7  or  3. 

Hence  the  number  is  73  or  37. 

The  two  values  of  x  are  the  required  digits  whose  sum  Is 
10.  It  will  be  observed  that  we  put  x  to  represent  the  first 
digit,  whereas  we  find  it  may  equal  the  second  as  well  as  the 
first.  The  reason  is,  that  we  have  here  imposed  a  condition 
which  does  not  enter  into  the  equation.  If  x  represent  either 
of  the  required  digits,  then  10—  a?  will  represent  the  other,  and 
hence  the  values  of  x  found  by  solving  the  equation  should 
give  both  digits.  Beginners  are  very  apt  thus,  in  the  state 
ment  of  a  problem,  to  impose  conditions  which  do  not  appear 
in  the  equation. 

The  preceding  example,  and  all  others  of  the  same  class, 
may  be  solved  without  completing  the  square.  Thus, 

Let  x  represent  the  half  difference  of  the  two  digits. 

Then,  according  to  the  principle  on  page  89,  5+  x  will  rep 
resent  the  greater  of  the  two  digits,  and  5—  #  the  less. 

The  square  of  5+o?  is  25  +  103?+  a?2, 
"  5-o?  "  25-10J7+  x2. 

The  sum  is  50  +2o?2,    which,   according   to 

the  problem,  =58. 

Hence  2x2=  8, 

or  oc2=  4, 

and  x  =±2. 

Therefore,  5+#  =7,  the  greater  digit, 

and  5—o?  =3,  the  less  digit. 

Prob.  8.  Find  two  numbers  such  that  the  product  of  their 
sum  and  difference  may  be  5,  and  the  product  of  the  sum  of 
their  squares  and  the  difference  of  their  squares  may  be  65. 


194  ALGEBRA. 

Prob.  9.  Find  two  numbers  such  that  the  product  of  their 
sum  and  difference  may  be  a,  and  the  product  of  the  sum  of 
their  squares  and  the  difference  of  their  squares  may  be  ma. 

Am. 

Prob.  10.  A  laborer  dug  two  trenches,  whose  united  length 
was  26  yards,  for  356  shillings,  and  the  digging  of  each  of 
them  cost  as  many  shillings  per  yard  as  there  were  yards  in 
its  length.  What  was  the  length  of  each  ? 

Ans.  10,  or  16  yards. 

Prob.  11.  What  two  numbers  are  those  whose  sum  is  2a, 
and  the  sum  of  their  squares  is  2b? 

Ans.  a+Vb  —  a'z,  and  a—Vb  —  a2. 

Prob.  12.  A  farmer  bought  a  number  of  sheep  for  80  dollars, 
and  if  he  had  bought  four  more  for  the  same  money,  he  would 
have  paid  one  dollar  less  for  each.  How  many  did  he  buy  ? 

Let  x  represent  the  number  of  sheep. 

80 

Then  will  —  be  the  price  of  each. 
x 

80 
And  — -  would  be  the  price  of  each  if  he  had  bought  four 

more  for  the  same  money. 
But  by  the  question  we  have 
80  _  80 
x  -SC4.4+1- 

Solving  this  equation,  we  obtain  #=16,  Ans. 

Prob.  13.  A  person  bought  a  number  of  articles  for  a  dol 
lars.  If  he  had  bought  2b  more  for  the  same  money,  he  would 
have  paid  c  dollars  less  for  each.  How  many  did  he  buy  ? 


Ans.  - 


Prob.  14.  It  is  required  to  find  three  numbers  such  that  the 
product  of  the  first  and  second  may  be  15,  the  product  of  the 
first  and  third  21,  and  the  sum  of  the  squares  of  the  second  and 
third  74.  Ans.  3,  5,  and  7. 


EQUATIONS  OF  THE  SECOND   DEGREE.  195 

Prob.  15.  It  is  required  to  find  three  numbers  such  that  the 
product  of  the  first  and  second  may  be  a,  the  product  of  the 
first  and  third  b,  and  the  sum  of  the  squares  of  the  second  and 

third  c.  ,  _  .  _  ,  _ 

/a2  +  b*  /     c          7     /     c 

Ans,  v/__;   a\/^^-    &V££&5- 

Prob.  16.  The  sum  of  two  numbers  is  16,  and  the  sum  of 
their  cubes  1072.  What  are  the  numbers?  Ans.  7  and  9. 

Prob.  17.  The  sum  of  two  numbers  is  2a,  and  the  sum  of 
their  cubes  is  2b.  What  are  the  numbers  ? 

Ans.  a  -f- 

Prob.  18.  Two  magnets,  whose  powers  of  attraction  are  as 
4  to  9,  are  placed  at  a  distance  of  20  inches  from  each  other. 
It  as  required  to  find,  on  the  line  which  joins  their  centres,  the 
point  where  a  needle  would  be  equally  attracted  by  both,  ad 
mitting,  that  the  intensity  of  magnetic  attraction  varies  inverse 
ly  as  the  square  of  the  distance. 

A        \  &  inches  from  the  weakest  magnet, 

(  or  —40  inches  from  the  weakest  magnet. 

Prob.  19.  Two  magnets,  whose  powers  are  as  m  to  ??,  are 
placed  at  a  distance  of  a  feet  from  each  other.  It  is  required 
to  find,  on  the  line  which  joins  their  centres,  the  point  which 
is  equally  attracted  by  both. 

The  distance  from  the  magnet  m  is  ' 


Ans. 

The  distance  from  the  magnet  n  is 


—. 

Vm  ±  V  n 

Prob.  20.  A  set  out  from  C  toward  D,  and  traveled  6  miles 
an  hour.  After  he  had  gone  45  miles,  B  set  out  from  D  to 
ward  C,  and  went  every  hour  -^V  of  the  entire  distance  ;  and 
after  he  had  traveled  as  many  hours  as  he  went  miles  in  one 
hour,  he  met  A.  Required  the  distance  between  the  places  C 
and  D.  Ans.  Either  100  miles,  or  180  miles. 


196  ALGEBRA. 

Prob.  21.  A  set  oat  from  C  toward  D,  and  traveled  a  miles 
per  hour.  After  he  had  gone  b  miles,  B  set  out  from  D  toward 
C,  and  went  every  hour  |th  of  the  entire  distance ;  and  after 
he  had  traveled  as  many  hours  as  he  went  miles  in  one  hour, 
he  met  A.  Kequired  the  distance  between  the  places  C  and  D. 


Prob.  22.  By  selling  my  horse  for  24  dollars,  I  lose  as  much 
per  cent,  as  the  horse  cost  me.  What  was  the  first  cost  of  the 
horse?  Ans.  40  or  60  dollars. 

Prob.  23.  A  fruit-dealer  receives  an  order  to  buy  18  melons 
provided  they  can  be  bought  at  18  cents  a  piece  ;  but  if  they 
should  be  dearer  or  cheaper  than  18  cents,  he  is  to  buy  as 
many  less  or  more  than  18  as  each  costs  more  or  less  than  18 
cents.  He  paid  in  all  $3.15.  How  many  melons  did  he  buy? 

Ans.  Either  15  or  21. 

Prob.  24.  A  line  of  given  length  (a)  is  bisected  and  pro 
duced  ;  find  the  length  of  the  produced  part,  so  that  the  rect 
angle  contained  by  half  the  line,  and  the  line  made  up  of  the 
half  and  the  produced  part,  may  be  equal  to  the  square  on  the 
produced  part. 


Equations  of  the  Second  Degree  containing  Two  Unknown 
Quantities. 

266.  An  equation  containing  two  unknown  quantities  is 
said  to  be  of  the  second  degree  when  the  highest  sum  of  the  expo 
nents  of  the  unknown  quantities  in  any  term  is  two.     Thus 

a?  +  ^  =  13,  (1.) 

and  x+xy+y  =  ll,  (2.) 

are  equations  of  the  second  degree. 

267.  The  solution  of  two  equations  of  the  second  degree 
containing  two  unknown  quantities  generally  involves  the  so 
lution  of  an  equation  of  the  fourth  degree  containing  one  un- 


EQUATIONS  OF  THE  SECOND  DEGREE.  197 

known  quantity.     Thus,  from  equation  (2),  we  find 

11-aj 

y=~z+r- 

Substituting  this  value  for  y  in  equation  (1)  and  reducing, 
we  have 


an  equation  which  can  not  be  solved  by  the  preceding  methods. 
Yet  there  are  particular  cases  in  which  simultaneous  equations 
of  a  degree  higher  than  the  first  may  be  solved  by  the  rules  for 
quadratic  equations.  The  following  are  the  principal  cases  of 
this  kind  : 

268.  1st.  When  one  of  the  equations  is  of  the  first  degree  and  the 
other  of  the  second.  —  We  find  an  expression  for  the  value  of  one 
of  the  unknown  quantities  in  the  former  equation,  and  substi 
tute  this  value  for  its  equal  in  the  other  equation. 


(x2  +  3xy-y2  =  23{  . 
Ex.  1.  Given  j  •  9  —  7        I  x         y' 

From  the  second  equation  we  find 


Substituting  this  value  for  x  in  the  first  equation,  we  have 
49-2%+42/2  +  21y-6/-2/2=:23? 

which  may  be  solved  in  the  usual  manner. 

(03=3  or  Q. 

Ans.  3  . 

=2  or  -- 


,,     0   n.  -  <_    .    ,          , 

Ex.  2.  Given  <  01        f  to  ^nd  x  and  y- 

[  ZX  —  o?/=:l 

(  x  =  5  or  — 
Ans.  J 


=     or  — 


(x  +  y_      j 

Ex.  3.  Given  •]  ~S~     ~~Xy  V  to  find  x  and  y. 
(9?/-9x=18) 


9?/ 

Ans. 


(  x=2  or  —4-. 

1 

(y=4  or  |. 

For  equations  of  this  class  there  are  in  general  two  sets  of 
values  of  x  and  y. 


198  ALGEBRA. 

269.  2d.  When  both  of  the  equations  are.  of  the  second  degree^ 
and  homogeneous. — Substitute  for  one  of  the  unknown  quanti 
ties  the  product  of  the  other  by  a  third  unknown  quantity. 

i/v»2    l    npji  —  1  0    / 
,  \  to  find  x  and  y. 
xy—2y2  =  L  ) 

If  we  assume  x=vy,  we  shall  have 

v2y2+vy2=~L2,  whence  y1  —  - 


12 
Therefore  -^— 


—       0. 

v2-hv    v—  2 

From  which  we  obtain      v=S  or  3. 

Substituting  either  of  these  values  in  one  of  the  preceding 
expressions  for  ?/2,  we  shall  obtain  the  values  of  y  ;  and  since 
x=vy,  we  may  easily  obtain  the  values  of  x. 


Ans.  _l 

y  =  ±  1  or  ±  —  r=. 

y  6 


Ex.  5.  Given 

—= 

Assuming  x—vy^  we  find   v=  j  or  —  . 


«^^-«.  x    «    , 

Ex.  6.  Given  |     5J2+^1:41      [  to  find  x  and  y. 

A  13 

Assuming  x=vy,  we  find  v  =  «  or 

o 

—  ±1  or  db 


=:db8  Or  ± 


EQUATIONS  OF  THE  SECOND  DEGREE.  199 

For  equations  of  this  class  there  are  in  general  four  sets  of 
values  of  x  and  y.  It  should  be  borne  in  mind  that  to  any 
one  of  the  four  values  of  x  there  corresponds  only  one  of  the 
four  values  of  y.  Thus,  when  x  in  the  6th  example  is  -f  1,  y 
must  be  -f  3,  and  can  not  be  one  of  the  other  three  values 
given  above. 

270.  3d.  When  the  unknown  quantities  enter  each  equation  sym 
metrically.  —  Substitute  for  the  unknown  quantities  the  sum  and 
difference  of  two  other  quantities,  or  the  sum  and  product  of 
two  other  quantities. 

/   r^l  nil  \ 

_j_^.-18 
Ex.  7.  Given  I  y     x          V  to  find  x  and  y. 

( 
Let  us  assume 


Then  x-\-y=2z-l2  or  z  =  6. 

That  is,  x—Q-\-v  and  y—Q—v. 

But  from  the  first  equation  we  find 


Substituting  the  preceding  values  of  x  and  y  in  this  equation, 
and  reducing,  we  have 


whence  v=±2. 

Therefore  x=4  or  8,  and  y—  8  or  4. 


T?      ,  n-  , 

Ex.  8.  Given  ]        y        0      >  to  find  x  and  y. 
(      x+y=8      ) 

lx=3  or  5. 
Ans.  \        K 

\y=5  or  3. 

Ex.  9.  Given  |   f  t^f  BQQA  \  ^  find  x  and  y. 
22=330) 


(  x=5  or  6. 
Ans.  1        a 

(  y—6  or  5. 

271.  4th.  When  the  same  algebraic  expression  is  involved  to 
different  powers,  it  is  sometimes  best  to  regard  this  expression 
as  the  unknown  quantity. 


200  ALGEBRA. 


Ex.10.  Given  j  -2+2-/+2/*+2*=120-22n  to  find  ,  and  y, 

(  xy  —  y  =  8  j 

The  first  equation  may  be  written 


Eegarding  x+y  as  a  single  quantity,  we  find  its  value  to  be 
either  10  or  —12. 

Proceeding  now  as  in  Art.  268,  we  find 
x=6  or  9,  or  —  9  +  ^5; 
y—4:  or  1,  or  —  3±V6. 


Ex.  11.  Given  -  to  find  x  and 

(      x+z/=6       j 

Eegarding  xy  as  the  unknown  quantity,  its  value  from  the 
first  equation  is  found  to  be  either 

8  or  -12. 

(x=2  or  4,  or  3±  V%1. 
Ans.  \  ,  _ 

(  y=4  or  2,  or  3  +  V21. 
<x*    4x_85) 

Ex.  12.  Given  •]  y2     y  ~  9   [to  find  x  and  y. 
(     x-y=2     ) 

Eegarding  -  as  the  unknown  quantity,  we  find  its  value  to 
be  either  5  17 

" 


For  several  of  these  examples  there  are  other  roots,  some 
of  which  can  not  be  obtained  by  the  processes  heretofore  ex 
plained.  The  roots  of  two  simultaneous  equations  are  some 
times  infinite,  as  in  the  equations  a?2—  xy  —  8,  and  a?2—?/2  =  12, 
since  two  quantities  that  are  infinitely  great  may  differ  \)j  a 
finite  quantity. 

Solve  the  following  groups  of  simultaneous  equations  : 


Ex.13. 


- 
Ans.  <       * 


EQUATIONS  OF  THE  SECOND  DEGREE. 

Ex.15.  =84>  ;         An  , 

=U) 


201 

=8or2. 
y=2or8. 


„ 
Ex' 


o:=6or30. 


V*  oo     (*5-2/5  =  3093| 
Ex'22'    <      s-t^S 


Ex  23 

j 

Ex.24.   ] 

( 

.  Put 


0^—5  or  —2. 
y=2  or  —5. 

J.ns.  ) 


2. 


PROBLEMS. 

1.  Divide  the  number  100  into  two  such  parts  that  the  sum 
of  their  square  roots  may  be  14.  Ans.  64  and  36. 

2.  Divide  the  number  a  into  two  such  parts  that  the  sum 
of  their  square  roots  may  be  b.  , 

A  ^   I   _    /o T% 

3.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their  fourth 

powers  is  706.     What  are  the  numbers  ?  Ans.  3  and  5. 

12 


202  ALGEBRA. 

4.  The  sum  of  two  numbers  is  2a,  and  the  sum  of  their 
fourth  powers  is  26.     What  are  the  numbers? 

Ans.  a±  V—Saz±V8a*+b. 

5.  The  sum  of  two  numbers  is  6,  and  the  sum  of  their  fifth 
powers  is  1056.     What  are  the  numbers?        Ans.  2  and  4. 

6.  The  sum  of  two  numbers  is  2a,  and  the  sum  of  their  fifth 
powers  is  b.     What  are  the  numbers  ? 


7.  What  two  numbers  are  those  whose  product  is  120  ;  and 
if  the  greater  be  increased  by  8  and  the  less  by  5,  the  product 
of  the  two  numbers  thus  obtained  shall  be  300  ? 

Ans.  12  and  10,  or  16  and  7.5. 

8.  What  two  numbers  are  those  whose  product  is  a;  and 
if  the  greater  be  increased  by  b  and  the  less  by  c,  the  product 
of  the  two  numbers  thus  obtained  shall  be  d? 


Ans.  Tr±V^i  ---  , 


,  d—a  —  bc 

where  m=  -    —  „ 
c 

9.  Find  two  numbers  such  that  their  sum,  their  product, 
and  the  difference  of  their  squares  may  be  all  equal  to  one 
another. 


that  is,  2.618,  and  1.618,  nearly. 

10.  Divide  the  number  100  into  two  such  parts  that  their 
product  may  be  equal  to  the  difference  of  their  squares. 

Ans.  38.197,  and  61.803. 

11.  Divide  the  number  a  into  two  such  parts  that  theii 
product  may  be  equal  to  the  difference  of  their  squares. 

3a±aV5        ,   —  a  +  aVE 
Ans.  -  -  and  -  -  • 


EQUATIONS  OF  THE   SECOND   DEGREE.  203 

12.  The  sum  of  two  numbers  is  a,  and  the  sum  of  their  re 
ciprocals  is  b.    Kequired  the  numbers. 


General  Properties  of  Equations  of  the  Second  Degree. 

272.  Every  equation  of  the  second  degree  containing  but  one 
unknown  quantity  has  two  roots,  and  only  two. 

We  have  seen,  Art.  250,  that  every  equation  of  the  second 
degree  containing  but  one  unknown  quantity  can  be  reduced 
to  the  form  x2+px  =  q.  We  have  also  found,  Art.  257,  that 
this  equation  has  two  roots,  viz., 


This  equation  can  not  have  more  than  two  roots  ;  for,  if  pos 
sible,  suppose  it  to  have  three  roots,  and  represent  these  roots 
by  a/,  x",  and  x'".  Then,  since  a  root  of  an  equation  is  such 
a  number  as,  substituted  for  the  unknown  quantity,  will  satisfy 
the  equation,  we  must  have 

q,  (1.) 

q,  (2.) 

=  q.  (3.) 

Subtracting  (2)  from  (1),  we  have 


Dividing  by  x'—x",  we  have 

(x'+x")+^=0-  (4.) 

In  the  same  manner,  we  find 

(a/+a/")+*=0.  (5.) 

Subtracting  (5)  from  (4),  we  have 
o;"-a;'"=:0; 

that  is,  the  third  supposed  root  is  identical  with  the  second  ; 
hence  there  can  not  be  three  different  roots  to  a  quadratic 
equation. 

273.   The  algebraic  sum  of  the  two  roots  is  equal  to  the  coeffi 
cient  of  the  second  term  of  the  equation  taken  with  the  contrary  sign. 


204  ALGEBRA. 

If  we  add  together  the  two  values  of  x  in  the  general  equa 
tion,  the  radical  parts  having  opposite  signs  disappear,  and  we 
obtain  p  p 

_  ±_   _  -*•      -    .        rf\ 

2      2~     p' 

Thus,  in  the  equation  x2—  I0x=—  16,  the  two  roots  are  8 
and  2,  whose  sum  is  +  10,  the  coefficient  of  x  taken  with  the 
contrary  sign. 

If  the  two  roots  are  equal  numerically,  but  have  opposite 
signs,  their  sum  is  zero,  and  the  second  term  of  the  equation 
vanishes.  Thus  the  two  roots  of  the  equation  xz—I6  are  +4 
and  —4,  whose  sum  is  zero.  This  equation  may  be  written 


274.  The  product  of  the  two  roots  is  equal  to  the  second  member 
of  the  equation  taken  with  the  contrary  sign. 

If  we  multiply  together  the  two  values  of  x  (observing  that 
the  product  of  the  sum  and  difference  of  two  quantities  is 
equal  to  the  difference  of  their  squares),  we  obtain 


Thus,  in  the  equation  x"2—  10x=  —  16,  the  product  of  the 
two  roots  8  and  2  is  +16,  which  is  equal  to  the  second  mem 
ber  of  the  equation  taken  with  the  contrary  sign. 

275.  The  last  two  principles  enable  us  to  form  an  equation 
whose  roots  shall  be  any  given  quantities. 

Ex.  1.  Find  the  equation  whose  roots  are  3  and  5. 

According  to  Art.  273,  the  coefficient  of  the  second  term  of 
the  equation  must  be  —  8  ;  and,  according  to  Art.  274,  the  sec 
ond  member  of  the  equation  must  be  —15.  Hence  the  equa 
tion  is  a;2—  8x=—  15. 

Ex.  2.  Find  the  equation  whose  roots  are  —4  and  —  7. 

Ex.  3.  Find  the  equation  whose  roots  are  5  and  —9. 

Ex.4.  Find  the  equation  whose  roots  are  —6  and  +11. 

Ex.  5.  Find  the  equation  whose  roots  are  1  and  —2. 

Ex.  6.  Find  the  equation  whose  roots  are  —  -J  and 

Ex.  7.  Find  the  equation  whose  roots  are  —  -J-  and 


EQUATIONS  OF  THE   SECOND  DEGREE.  205 

Ex.  8.  Find  the  equation  whose  roots  are  1  ±  \/5. 
Ex.  9.  Find  the  equation  whose  roots  are  1  ±  V^5. 

276.  Every  equation  of  the  second  degree  whose  roots  are  a  and 
b,may  be  reduced  to  the  form  (x—a)(x—b)  —  0. 
Take  the  general  equation 


and  write  it  x2+px—  q  —  0. 

Then,  by  Art.  273,       p=-(a  +  b); 
and  by  Art.  274,  q=  —ab. 

Hence,  by  substitution, 


or,  resolving  into  factors, 

(x  —  a)(x-b)  =  Q. 

Thus  the  equation  x2—  10x=—  16,  whose  roots  are  8  and  2, 
may  be  resolved  into  the  factors  x  —  8  =  0  and  x—  2  =  0. 

It  is  also  obvious  that  if  a  is  a  root  of  an  equation  of  the 
second  degree,  the  equation  must  be  divisible  by  x—  a.  Thus 
the  preceding  equation  is  divisible  by  x—  8,  giving  the  quo 
tient  x—2. 

Ex.  1.  The  roots  of  the  equation  x2-f  6x4-8  =  0  are  —2  and 
—  4.  Eesolve  the  first  member  into  its  factors. 

Ex.  2.  The  roots  of  the  equation  x24-  6x—  27  =  0  are  +3 
and  —9.  Resolve  the  first  member  into  its  factors. 

Ex.  3.  The  roots  of  the  equation  x2  —  2x—  24  =  0  are  +6 
and  —4.  Resolve  the  first  member  into  its  factors. 

Ex.4.  Resolve  the  equation  x24-73x+7SO  =  0  into  simple 
factors.  Ans.  (x  +  60)  (x  +  13)  =  0. 

Ex.5.  Resolve  the  equation  x2—  88x4-1612  =  0  into  simple 
factors.  Ans.  (x-62)(x-26)  =  0. 

Ex.6.  Resolve  the  equation  2x2+x—  6  =  0  into  simple  fac 
tors.  Ans.  2(x+2)0-!)  =  0. 

Ex.7.  Resolve  the  equation  3x2—  Wx—  25  =  0  into  simple 
Actors.  A  ns.  3  (x  —  5)  (x  +  f  )  =  0. 


206  ALGEBRA. 

Discussion  of  the  General  Equation  of  the  Second  Degree. 

277.  In  the  general  equation  of  the  second  degree  x2-{-px—q. 
the  coefficient  of  x,  as  well  as  the  absolute  term,  may  be  either 
positive  or  negative.  We  may  therefore  have  the  four  fol 
lowing  forms: 

First  form,  x2 -\-px- q. 

Second  form,  x2—px  —  q. 

Third  form,  x2 -\-px-  —  q. 

Fourth  form,  x2  —px  ——q. 

From  these  equations  we  obtain 


4 

We  will  now  consider  what  conditions  will  render  these 
roots  positive  or  negative,  equal  or  unequal,  real  or  imaginary. 

278.  Positive  and  negative  roots. 

Since  "A"^  *s  &reater  tnan  ~n 

y  4- +2  must  be  greater  than  If- 
For  the  same  reason,  y  ~  —  q  must  be  less  than  ^. 

Therefore,  in  the  first  and  second  forms,  the  sign  of  the  roots 
will  correspond  to  the  sign  of  the  radicals;  but  in  the  third 
and  fourth  forms  the  sign  of  the  roots  will  correspond  to  the 
sign  of  the  rational  parts.  Hence,  in  the  first  form,  one  root  is 
positive  and  the  other  negative,  and  the  negative  root  is  numerical 
ly  the  greatest;  as  in  the  equation  x2+x=6,  whose  roots  are 
-4-2  and  —3. 


EQUATIONS   OF  THE   SECOND   DEGEEE.  207 

In  the  second  form  one  root  is  positive  and  the  other  negative, 
and  the  positive  root  is  numerically  the  greatest,  as  in  the  equa 
tion  x2— x  —  Q,  whose  roots  are  —2  and  +3. 

In  the  third  form  both  roots  are  negative,  as  in  the  equation 
xz-\-6x=—  6,  whose  roots  are  —2  and  —3. 

In  the  fourth  form  both  roots  are  positive,  as  in  the  equation. 
<c2— 5x=—  6,  whose  roots  are  +2  and  +3. 

279.  Equal  and  unequal  roots. 

In  the  first  and  second  forms  the  two  roots  are  always  unequal; 
but  in  the  third  and  fourth  forms,  when  q  is  numerically  equal 

to  — ,  the  radical  part  of  both  values  of  x  becomes  zero,  and 
4 

the  two  roots  are  then  said  to  be  equal     In  this  case 

7)  T) 

the  third  form  gives   x=  —  ~±0=  —  £ 
and  the  fourth  form  gives 


Thus,  in  the  equation  x2  +  6x=  —  9,  the  two  roots  are  —3 
and  —3.  We  say  that  in  this  case  the  equation  has  two 
roots,  because  it  is  the  product  of  the  two  factors  03+8  =  0 
and  x  +  3  =  0. 

So,  also,  in  the  equation  x2— 6x=  — 9,  the  two  roots  are  -f  3 
and  +3. 

280.  Real  and  imaginary  roots. 

rpT. 
Since  ^-,  being  a  square,  is  positive  for  all  real  values  of  p, 

p2 

it  follows  that  the  expression  ±-  -{-q  can  only  be  rendered  neg 
ative  by  the  sign  of  q ;  that  is,  the  quantity  under  the  radical 
sign  can  only  be  negative  when  q  is  negative  and  numerically 

TJ2 

greater  than  -—.  Hence,  in  the  first  and  second  forms,  both  roots 
are  always  real;  but  in  the  third  and  fourth  forms  both  roots  arc 
imaginary  when  q  is  numerically  greater  than  — . 


208  ALGEBRA. 

Thus,  in  the  equation  £2+4x=  —  6,  the  two  roots  are 
and  in  the  equation  x2— 4x=  —  6,  the  two  roots  are 

It  will  be  observed  that  when  one  of  the  roots  is  imaginary, 
the  other  is  imaginary  also. 

281.  Imaginary  roots  indicate  impossible  conditions  in  the  pro 
posed  question  which  furnished  the  equation. 

The  demonstration  of  this  principle  depends  upon  the  fol 
lowing  proposition :  the  greatest  product  which  can  be  obtained 
by  dividing  a  number  into  two  parts  and  multiplying  them  togeth' 
er  is  the  square  of  half  that  number. 

Let  p  represent  the  given  number,  and  d  the  difference  of 
the  parts.  , 

Then,  from  page  89,  |?+-  =  the  greater  part, 

T  P     d      ,     , 

and  ^-—-  =  the  less  part, 

and  ^ —  —  =  the  product  of  the  parts. 

Now,  since  p  is  a  given  quantity,  it  is  plain  that  the  prod 
uct  will  be  the  greatest  possible  when  c?=0;  that  is,  the  great- 

JD 

est  product  is  the  square  of  ^-,  half  the  given  number. 

For  example,  let  10  be  the  number  to  be  divided. 
We  have  10  =  1  +  9;    and  9x1=   9. 

10  =  2  +  8;  and  8x2  =  16. 
10  =  3  +  7;  and  7x3  =  21. 
10=4  +  6;  and  6x4  =  24. 
10=5  +  5;  and  5x5  =  25. 

Thus  we  see  that  the  smaller  the  difference  of  the  two  parts, 
the  greater  is  their  product ;  and  this  product  is  greatest  when 
the  two  parts  are  equal. 

Now,  in  the  equation  x2—px  = — j,  p  is  the  sum  of  the  two 
roots,  and  q  is  their  product.  Therefore  q  can  never  be  great 
er  than  •£-. 


EQUATIONS   OF  THE   SECOND  DEGREE.  209 

If,  then,  any  problem  furnishes  an  equation  in  which  q  is 

p2 
negative,  and  numerically  greater  than  ^j-,  we  infer  that  the 

conditions  of  the  question  are  incompatible  with  each  other. 

Suppose  it  is  required  to  divide  6  into  two  parts  such  that 
their  product  shall  be  10. 

Let  x  represent  one  of  the  parts,  and  6  —  x  the  other  part 

Then,  by  the  conditions, 

x(Q-x)  =  W; 

whence  '   x2— 6x=— 10, 

and  x  =  3±V^l. 

The  imaginary  value  of  x  indicates  that  it  is  impossible  to 
find  two  numbers  whose  sum  is  6  and  product  10.  From  the 
preceding  proposition,  it  appears  that  9  is  the  greatest  product 
which  can  be  obtained  by  dividing  6  into  two  parts  and  multi 
plying  them  together. 

Discussion  of  Particular  Problems. 

282.  In  discussing  particular  problems  which  involve  equa 
tions  of  the  second  degree,  we  meet  with  all  the  different  cases 
which  are  presented  by  equations  of  the  first  degree,  and  some 
peculiarities  besides.     All  the  different  cases  enumerated  in 
Chapter  X.  are   presented  by  Prob.  19,  page  195,  when  we 
make  different  suppositions  upon  the  values  of  a,  m,  and  n; 
but  we  need  not  dwell  upon  them. 

The  peculiarities  exhibited  by  equations  of  the  second  de 
gree  are  double  values  of  x,  and  imaginary  values. 

283.  Double  Values  of  the  Unknown  Quantity. — We  have  seen 
that  every  equation  of  the  second  degree  has  two  roots.    Some 
times  both  of  these  values  are  applicable  to  the  problem  which 
furnishes  the  equation.    Thus,  in  Prob.  20,  page  195,  we  obtain 
either  100  or  180  miles  for  the  distance  between  the  places  C 

and  D. 

C  E  D 


Let  E  represent  the  position  of  A  when  B  sets  out  on  his 


210  ALGEBRA. 

journey.  Then,  if  we  suppose  CD  equals  100  miles,  ED  will 
equal  55  miles,  of  which  A  will  travel  30  miles  (being  6  miles 
an  hour  for  5  hours),  and  B  will  travel  25  miles  (being  5  miles 
an  hour  for  5  hours). 

If  we  suppose  CD  equals  180  miles,  ED  will  equal  135  miles, 
of  which  A  will  travel  54  miles  (being  6  miles  an  hour  for  9 
hours),  and  B  will  travel  81  miles  (being  9  miles  an  hour  for 
9  hours). 

This  problem,  therefore,  admits  of  two  positive  answers,  both 
equally  applicable  to  the  question.  Problems  22  and  23,  page 
196,  are  of  the  same  kind. 

In  Problem  18,  page  195,  one  of  the  values  of  x  is  positive 
and  the  other  negative. 

C'  A  C  B 


Let  the  weaker  magnet  be  placed  at  A,  and  the  stronger  at 
B ;  then  C  will  represent  the  position  of  a  needle  equally  at 
tracted  by  both  magnets.  According  to  the  first  value,  the 
distance  AC =8  inches,  and  CB  =  12  inches.  Now,  at  the  dis 
tance  of  8  inches,  the  attraction  of  the  weaker  magnet  will  be 

4  * 
represented  by  —2 ;    and  at  the  distance  of  12  inches,  the  at- 

9 

traction  of  the  other  magnet  will  be  represented  by  ir^,  and 

these  two  powers  are  equal ;  for 

1-JL 

S2~122' 

But  there  is  another  point,  C',  which  equally  satisfies  the 
conditions  of  the  question ;  and  this  point  is  40  inches  to  the 
left  of  A,  and  therefore  60  inches  to  the  left  of  B ;  for 

A     JL 

402~602' 

284.  Imaginary  Values  of  the  Unknown  Quantity. — We  have 
seen  that  an  imaginary  root  indicates  impossible  conditions  in 
the  proposed  question  which  furnished  the  equation.  In  sev 
eral  of  the  preceding  problems  the  values  of  x  become  imag 
inary  in  particular  cases. 


EQUATIONS  OF  THE  SECOND  DEGREE.  211 

When  will  the  values  of  x  in  Prob.  6,  page  192,  be  imag 
inary  ?  Ans.  When  b  >  a2. 

What  is  the  impossibility  involved  in  this  supposition  ? 

Ans.  It  is  impossible  that  the  product  of  two  numbers  can 
be  greater  than  the  square  of  half  their  sum. 

When  will  the  values  of  x  in  Prob.  11,  page  194,  be  imag 
inary  ?  Ans.  When  a2  >  b  ;  or  (2a)2  >  46. 

What  is  the  impossibility  involved  in  this  supposition? 

Ans.  The  square  of  the  sum  of  two  numbers  can  not  be 
greater  than  twice  the  sum  of  their  squares. 

When  will  the  values  of  x  in  Prob.  17,  page  195,  be  imag 
inary  ?  Ans.  When  a3  >  b  ;  or  (2a)3  >  85. 

What  is  the  impossibility  of  this  supposition  ? 

Ans.  The  cube  of  the  sum  of  two  numbers  can  not  be  great 
er  than  four  times  the  sum  of  their  cubes. 

When  will  the  values  of  x  in  Prob.  4,  page  180,  be  imag 
inary,  and  what  is  the  impossibility  of  this  supposition? 

285.  Geometrical  Construction  of  Equations  of  the  Second  De 
gree.  —  The  roots  of  an  equation  of  the  second  degree  may  be 
represented  by  a  simple  geometrical  figure.  This  may  be  done 
for  each  of  the  four  forms  : 

First  form.  —  The  first  form  gives  for  x  the  two  values 
TJ         f~r?  v 

Xz=~f  +  V  T+?>  a      *=-f--Y.f 
Draw  the  line  AB,  and  make  it  equal  to  -^fq.    From  B  draw 
BO  perpendicular  to  AB,  and  make  it  equal 

to  ~r.    Join  A  and  C  ;  then  AC  will  repre- 

j-i  .  - 

sent  the  value  of       ¥-+.     For 


B  AB2+BC2  (Geom.,  Prop.  11,  Bk.  IY.). 

With  C  as  a  centre,  and  CB  as  a  radius,  describe  a  circle 
cutting  AC  in  D,  and  AC  produced  in  E.  For  the  first  value 
of  x  the  radical  is  positive,  and  is  set  off  from  A  toward  C  ; 

then  —  |J  is  set  off  from  C  to  D  ;   and  AD.  estimated  from  A 
2i 

to  D,  represents  the  first  value  of  x. 


212  ALGEBRA. 

For  the  second  value  of  x  we  begin  at  E,  and  set  off  EC 

/y-\ 

equal  to  —  |j  ;  we  then  set  off  the  minus  radical  from  C  to  A  j 

and  EA,  estimated  from  E  to  A,  represents  the  second  value 
of  x. 

Second  form. — The  second  form  gives  for  x  the  two  values 


The  first  value  of  x  is  represented  by  AE  estimated  from  A 
to  E.  The  second  value  is  -f  DC  —  CA,  the  latter  being  esti 
mated  from  C  to  A.  Hence  the  second  value  is  represented 
by  DA  estimated  from  D  to  A. 

TJiirdform. — The  third  form  gives  for  x  the  two  values 


*T-g,  and  »=~|. 

Draw  an  indefinite  line  FA,  and  from  any  point,  as  A,  set 
0  off  a  distance  AB=— J?.      We  set  off  its 

value  to  the  left,  because  £  is  negative. 
JF  D 'E  A  At  B  draw  BC  perpendicular  to  FA,  and 
make  it  equal  to  Vq.  With  C  as  a  centre,  and  a  radius  equal 

to  -^,  describe  an  arc  of  a  circle  cutting  FA  in  D  and  E.    Now 
A  , 

the  value  of  y  ^-  —  q  will  be  BD  or  BE.     The  first  value  of 

x  will  be  represented  by  —  AB  +  BE,  which  is  equal  to  —  AE. 
The  second  will  be  represented  by  — AB  —  BD,  which  is  equal 
to  — AD;  so  that  both  of  the  roots  are  negative,  and  are  esti 
mated  in  the  same  direction,  from  A  toward  the  left. 

Fourth  form. — The  fourth  form  gives  for  x  the  two  values 


T~?'  and  x=~y  4  ~q- 

Construct  the  radical  part  of  the  value  of  x  as  in  the  last 

79 

~ 

a 


79 

case.     Then,  since  ~  is  positive,  we  set  off  its  value  AB  from 


EQUATIONS  OF  THE   SECOND  DEGREE.  213 

A  toward  the  right.  To  AB  we  add  BD, 
which  gives  AD  for  the  first  value  of  x; 
and  from  AB  we  subtract  BE,  which  leaves 
AE  for  the  second  value  of  x.  Both  val 
ues  are  positive,  and  are  estimated  in  the 
same  direction,  from  A  toward  the  right. 

Equal  Roots.— If  the  radius  CE  be  taken  equal  to  CB,  that 

is,  if  Vq  is  equal  to  -f,  the  arc  described  with  the  centre  C  will 

A 

be  tangent  to  AF,  the  two  points  D  and  E  will  unite,  and  the 
two  values  of  x  become  equal  to  each  other.  In  this  case  the 
radical  part  of  the  value  of  x  becomes  zero. 

Imaginary  Boots. — If  the  radius  of  the  circle  described  with 
the  centre  C  be  taken  less  than  CB,  it  will  not 


/      meet  the  line  AF.    In  this  case  q  is  numerical- 

2 

ly  greater  than  ±L,  and  the  radical  part  of  the 


value  of  x  is  imaginary. 


214  ALGEBRA. 


CHAPTER  XV. 

RATIO    AND    PROPORTION. 

286.  Ratio  is  the  relation  which  one  quantity  bears  to  an 
other  with  respect  to  magnitude.      Ratio  is  denoted  by  two 
points  like  the  colon  (:)  placed  between  the  quantities  com 
pared.     Thus  the  ratio  of  a  to  b  is  written  a :  b. 

The  first  quantity  is  called  the  antecedent  of  the  ratio,  and 
the  second  the  consequent.  The  two  quantities  compared  are 
called  the  terms  of  the  ratio,  and  together  they  form  a  couplet. 
The  quantities  compared  may  be  polynomials;  nevertheless, 
each  quantity  is  called  one  term  of  the  ratio. 

287.  A  ratio  is  measured  by  the  fraction  whose  numerator 
is  the  antecedent  and  whose  denominator  is  the  consequent  of 

the  ratio.     Thus  the  ratio  of  a  to  b  is  measured  by  j. 

288.  A  compound  ratio  is  the  ratio  arising  from  multiplying 
together  the  corresponding  terms  of  two  or  more  simple  ratios. 
Thus  the  ratio  of  a  to  b  compounded  with  the  ratio  of  c  to  d 
becomes  ac  to  bd. 

The  ratio  compounded  of  the  ratios  3  to  5  and  7  to  9  is 
21  to  45. 

289.  The  duplicate  ratio  of  two  quantities  is  the  ratio  of 
their  squares.     Thus  the  duplicate  ratio  of  2  to  3  is  4  to  9 ; 
the  duplicate  ratio  of  a  to  b  is  a2  to  b2. 

290.  The  triplicate  ratio  of  two  quantities  is  the  ratio  of  their 
cubes.     Thus  the  triplicate  ratio  of  a  to  b  is  a3  to  b3. 

291.  If  the  terms  of  a  ratio  are  both  multiplied  or  both  divided 


RATIO  AND  PROPORTION.  215 

by  the  same  quantity,  the  value  of  the  ratio  remains  unchanged. 
The  ratio  of  a  to  b  is  represented  by  the  fraction  j,  and  the 

value  of  a  fraction  is  not  changed  if  we  multiply  or  divide 
both  numerator  and  denominator  by  the  same  quantity.  Thus 

a    ma 

b=mb> 

7  7     a    b 

or  a :  b=.ma  :  mb— - :  -. 

.;./:.  n   n 

PROPORTION. 

292.  Proportion  is  an  equality  of  ratios.     Thus,  if  a,  b,  c,  d 
are  four  quantities  such  that  a  when  divided  by  b  gives  the 
same  quotient  as  c  when  divided  by  d,  these  four  quantities 
are  called  proportionals.    This  proportion  may  be  written  thus, 

a :  b  : :  c  :  d, 
or  a:b=.c:d, 

°r  f=> 

b      d 
In  either  case  the  proportion  is  read  a  is  to  b  as  c  is  to  d. 

293.  The  terms  of  a  proportion  are  the  four  quantities  which 
are  compared.     The  first  and  fourth  terms  are  called  the  ex 
tremes,  the  second  and  third  the  means.    The  first  term  is  called 
the  first  antecedent,  the  second  term  the  first  consequent,  the  third 
term  the  second  antecedent,  and  the  fourth  term  the  second  con 
sequent. 

294.  When  the  first  of  a  series  of  quantities  has  the  same 
ratio  to  the  second  that  the  second  has  to  the  third,  or  the 
third  to  the  fourth,  and  so  on,  these  quantities  are  said  to  be  in 
continued  proportion,  and  any  one  of  them  is  a  mean  propor 
tional  between  the  two  adjacent  ones.     Thus,  if 

a :  b : :  b :  c  : :  c :  d : :  d :  e, 

then  a,  b,  c,  d,  and  e  are  in  continued  proportion,  and  b  is  a 
mean  proportional  between  a  and  c,  c  is  a  mean  proportional 
between  b  and  d,  and  so  on. 


216  ALGEBRA. 

295.  Alternation  is  when  antecedent  is  compared  with  ante 
cedent  and  consequent  with  consequent.     Thus,  if 

a  :  b  : :  c  :  d, 
then,  by  alternation,          a  :  c  : :  b  :  d.     See  Art.  301. 

296.  Inversion  is  when  antecedents  are  made  consequents, 
and  consequents  are  made  antecedents.     Thus,  if 

a:b::c:d) 
then,  inversely,  b:  a::  d:  c.     See  Art.  302. 

297.  Composition  is  when  the  sum  of  antecedent  and  conse 
quent   is    compared   with   either    antecedent  or   consequent. 
Thus,  if  a    b::c:d, 

then,  by  composition,  a-+  b  :  a  : :  c+d :  c, 

and  a-\-b:b::  c+d :  d.     See  Art.  304. 

298.  Division  is  when  the  difference  of  antecedent  and  con 
sequent  is  compared  with  either  antecedent  or  consequent. 
Thus,  if  a:b::c:d, 

then,  by  division,         a- -b  :  a  : :  c— d :  c, 

and  a— b  :b::  c—d :  d.     See  Art.  305. 

299.  If  four  quantities  are  in  proportion,  the  product  of  the  ex 
tremes  is  equal  to  the  product  of  the  means. 

Let  a:b  ::  c:  d. 

Then  |  =  J,  Art.  292. 

Multiplying  each  of  these  equals  by  bd,  we  have  ad—lc. 

300.  Conversely,  if  the  product  of  tiuo  quantities  is  equal  to  the 
product  of  two  other  quantities,  the  first  two  may  be  made  the  ex 
tremes,  and  the  other  two  the  means  of  a  proportion. 

Let  ad—lc. 

Dividing  each  of  these  equals  by  bd,  we  have 

a_  £ 

b~d? 
or  a:b::c:d,  Art.  292. 


RATIO  AND  PROPORTION.  217 

EXAMPLES. 

1.  Given  the  first  three  terms  of  a  proportion,  24,  15,  and 
40,  to  find  the  fourth  term. 

2.  Given  the  first  three  terms  of  a  proportion,  3a&3,  4a262, 
and  9a3b,  to  find  the  fourth  term. 

3.  Given  the  last  three  terms  of  a  proportion,  4a365,  3a363, 
and  2a5£,  to  find  the  first  term. 

4.  Given  the  first,  second,  and  fourth  terms  of  a  proportion, 
5/,  7x2#3,  and  2lxGy,  to  find  the  third  term. 

5.  Given  the  first,  third,  and  fourth  terms  of  a  proportion, 
a  +5,  a2—  &2,  and  (a  —  &)2,  to  find  the  second  term. 

Which  of  the  following  proportions  are  correct,  and  which 
are  incorrect? 

6.  3a+46:  9a  +  86::  a-2b:  Ba—  4b. 

7.  9a2-4£2  :  I5a2-25ab+8b2  :  :  15a2+25a6+852:  25a2-1662. 
8. 

9. 


301.  If  four  quantities  are  in  proportion,  they  will  be  in  pro 
portion  when  taken  alternately. 

Let  a  :  b  ::  c:  d; 

then 

Multiplying  by  6, 

Dividing  bye,  ~c=\> 

or  a:c::b:d. 

302.  If  four  quantities  are  in  proportion,  they  ivill  be  in  pro 
portion  wJien  taken  inversely. 

Let  a:b::c:d; 

then  £  =  £ 

6      a 

Divide  unity  by  each  of  these  equal  quantities,  and  we  have 

b_d 
a~  c' 

or  b  :  a  :  :  d  :  c. 

K 


218  ALGEBRA. 

303.  fiatios  tiial  are  equal  to  the  same  ratio  are  equal  to  each 
other. 

If  albumin,                          (1.) 

and  c :  d : :  m :  «,                          (2.) 

then  a  :  b : :  c :  d 

From  proportion  (1),  y  =  — . 

From  proportion  (2)^  -^  =  — . 

TT  a       c 

Hence  1  =d> 

or  aibiicid. 


304.  //"  /cwr  quantities  are  proportional,  they  will  be  propor 
tional  by  composition. 

Let  aibiicid; 

,  a       c 

then  j-  =  -3. 

Add  unity  to  each  of  these  equals,  and  we  have 


.    , .                                   a+b      c+d 
that  is,  = , 

or  a  +  b  i  bn  c-frf:  d. 

305.  If  four  quantities  are  proportional,  they  will  be  propor 
tional  by  division. 

L  :  a  ib  nc  id; 

then  |  =  |. 

Subtract  unity  from  each  of  these  equals,  and  we  have 

a     1  —  c     1 
b~~     ~d~ 

,       .  a  — ^      c— c? 

that  is,  — T—  =  — -i— , 

or  a  —  b-b:ic—did. 


RATIO   AND  PROPORTION.  219 

306.  If  four  quantities  are  proportional,  the  sum  of  the  first 
and  second  is  to  vieir  difference^as  the  sum  of  the  third  and  fourth 
is  to  their  difference. 

L  -.-:  a:b::c:d. 

By  composition.  Art.  304^ 

a+b:b::c+d:d. 
By  alternation,  Art  301, 

a+b:c+d::b:d. 
Also  by  division,  Art.  305, 

a  —  b:  b::  c—d'.d; 

by  alternation,  a—  b  :  c—d  ::b:d. 

By  equality  of  ratios,  Art  303, 

a+bi  c+d'.i  a—b:  c  —  d, 
or  a-r-b:  a—b::  c+d:  c—  d. 

307.  If  four  quantities  are  in  proportion,  any  equimultiples  of 
the  first  couplet  will  be  proportional  to  any  equimultiples  of  the 
second  coupkL 

Let  a:b::c:d; 

***  f-3-  :'"..• 

Multiply  both  terms  of  the  first  fraction  by  w»,  and  both 
terms  of  the  second  fraction  by  w,  and  we  have 

:  _       : 
.  : 
or  maimb-.-ncind, 

303.  If  four  quantities  are  in  proportion,  any  equimultiples  of 
the  antecedents  wiO.  be  proportional  to  any  equimultiples  of  the  con- 


L  I  a:b::c:d; 

then  a  _  c 

b~d' 

Multiply  each  of  these  equals  by  m,  and  we  have 
ma      me 


220  ALGEBRA. 

Divide  each  of  these  equals  by  w, 
ma      me 
nb  ~~  nd> 
or  ma  :nb::mc:  rid. 

309.  If  any  number  of  quantities  are  proportional,  any  one  an 
tecedent  is  to  its  consequent  as  the  sum  of  all  the  antecedents  is  to 
the  sum  of  all  the  consequents. 

Let  a  :  b  :  :  c  :  d  :  :  e  :  /; 

then,  since  a  :  b  :  :  c  :  d, 

ad  =  bc;  (1.) 

and,  since  a  :  b  :  :  e  :  /, 

«/=**/  (2-) 

also  ab  =  ba.  (3.) 

Adding  (1),  (2),  and  (3), 

ab+ad+af  =  ba+bc+be; 
that  is,  a  (b  +  d+f)  =  b(a  +  c  +  e). 

Hence,  Art.  300,  a  :  b  ::  a+c  +  e  :  b  +  d+f 

310,  If  there  are  two  sets  of  proportional  quantities,  the  prod 
ucts  of  the  corresponding  terms  will  be  proportional. 

Let  a  :  b  :  :  c  :  d, 

and  e:f::g:h; 

then  ae  :  bf  :  :  eg  :  dh. 

- 


Multiplying  together  these  equal  quantities,  we  have 

ae_cg_ 
bf  ~  dh1 
or  ae  :bf::  eg  :  dh. 

311.  If  four  quantities  are  in  proportion,  like  powers  or  roots 
of  these  quantities  will  also  be  in  proportion. 


RATIO   AND   PKOPOKTION. 


221 


a:b::c:d; 

a      c 

T  —  -j- 
b      d 

Eaising  each  of  these  equals  to  the  nth  power,  we  obtain 


Let 
then 


that  is,  an:bn  ::  cn:  dn. 

In  the  same  manner,  we  find 

i      i.       i      i 
an :  bn  ::  cn :  dn. 

312.  If  three  quantities  are  in  continued  proportion,  the  product 
of  the  extremes  is  equal  to  the  square  of  the  mean. 

If  a:b::b:c, 

then,  by  Art.  299,  ac=bb=b2. 

313.  If  three  quantities  are  in  continued  proportion,  the  first  is 
to  the  third  in  the  duplicate  ratio  of  the  first  to  the  second. 

Let  a:b::b:c; 

a  _  b 
b~  c 
Multiply  each  of  these  equals  by  y,  and  we  have 

a     a     a     b 
bXb=bXc; 

a_a? 

c      62' 
a  :  c  : :  a2 :  b2. 


then 


that  is, 
or 


314.  If  four  quantities  are  in  continued  proportion,  the  first  is 
to  the  fourth  in  the  triplicate  ratio  of  the  first  to  the  second. 
Let  a:b  ::b  :  c::  c:  d ; 

a      b 

be 

a  _  c 

b~d' 

a  _  a 

b=b' 


then 

and 

also 


(2.) 
(3.) 


222  ALGEBRA. 

Multiplying  together  (1),  (2),  and  (3),  we  have 
a3      dbc      a 


nence  a  :  d  :  :  a3  :  b3. 

VARIATION. 

315.  Proportions  are  often  expressed  in  an  abridged  form. 
Thus,  if  A  and  B  represent  two  sums  of  money  put  out  for 
one  year  at  the  same  rate  of  interest,  then 

A  :  B  :  :  interest  of  A  :  interest  of  B. 

This  is  briefly  expressed  by  saying  that  the  interest  varies  as 
the  principal.  A  peculiar  character  (  a  )  is  used  to  denote  this 
relation.  1  hus  interest  oc  principal 

denotes  that  the  interest  varies  as  the  principal. 

316.  One  quantity  is  said  to  vary  directly  as  another  when 
the  two  quantities  increase  or  decrease  together  in  the  same 
ratio.     Thus,  in  the  above  example,  A  varies  directly  as  the 
interest  of  A.     In  such  a  case,  either  quantity  is  equal  to  the 
other  multiplied  by  some  constant  number. 

Thus,  if  the  interest  varies  as  the  principal,  then  the  interest 
equals  the  product  of  the  principal  by  some  constant  number, 
which  is  the  rate  of  interest. 

If  A  a  B,  then  A  =  raB. 

If  the  space  (S)  described  by  a  falling  body  varies  as  the 
square  of  the  time  (T),  then 

S  =  mT2, 
where  m  represents  a  constant  multiplier. 

317.  One  quantity  may  vary  directly  as  the  product  of  sev 
eral  others. 

Thus,  if  a  body  moves  with  uniform  velocity,  the  space  de 
scribed  is  measured  by  the  product  of  the  time  by  the  velocity. 
If  we  put  S  to  represent  the  space  described,  T  the  time  of 
motion,  and  V  the  uniform  velocity,  then  we  shall  have 

S  a  T  x  V. 


RATIO  AND  PROPORTION.  223 

Also  the  area  of  a  rectangular  figure  varies  as  the  product 
of  its  length  and  breadth. 

The  weight  of  a  stick  of  timber  varies  as  the  product  of  its 
length  x  its  breadth  x  its  depth  x  its  density. 

318.  One  quantity  is  said  to  vary  inversely  as  another  when 
the  first  varies  as  the  reciprocal  of  the  second.     Thus,  if  the 
area  of  a  triangle  be  invariable,  the  altitude  varies  inversely  as 
the  base. 

If  the  product  of  two  quantities  is  constant,  then  one  varies 
inversely  as  the  other. 

In  uniform  motion  the  space  described  is  measured  by  the 
product  of  the  time  by  the  velocity ;  that  is, 

SocTxV; 

S 
whence  T  a  =. 

If  the  space  be  supposed  to  remain  constant,  then 

Toe  —  • 

y 

that  is,  the  time  required  to  travel  a  given  distance  varies  in 
versely  as  the  velocity. 

Conversely,  if  one  quantity  varies  inversely  as  another,  the 
product  of  the  two  quantities  is  constant 

Thus,  if  T  a  i 

then  the  product  of  T  by  Y  is  equal  to  a  constant  quantity. 

319.  One  quantity  is  said  to  vary  directly  as  a  second,  and 
inversely  as  a  third,  when  it  varies  as  the  product  of  the  second 
by  the  reciprocal  of  the  third.     Thus,  according  to  the  New 
tonian  law  of  gravitation,  the  attraction  (G)  of  any  heavenly 
body  varies  directly  as  the  quantity  of  matter  (Q),  and  inverse 
ly  as  the  square  of  the  distance  (D). 

That  is,  Gr  a      . 


224  ALGEBRA. 

EXAMPLES. 

~.         |  x+y:x::  5:3)  , 

1.  Given  <  „         f  to  find  x  and  y. 

(         xy=6          } 

Since  x+y:  x  ::  5  :  3, 

by  division,  Art.  305,       y  :  x  :  :  2  :  3. 

2x 
Hence  3z/=2x,  and  y=-~-- 

Substituting  this  value  of  y  in  the  second  equation,  we  ob 
tain  %X2 

T-: 

Therefore  x=±S, 

and  y  =  ±2. 

~.          (  x  +  y  :  x—  y  ::  3  :  1  }        «    ,  , 

2.  Given  i        *,       »     k/i        f  to  find  x  and  y. 

(       ^_  ?/3  —  56        ) 

From  the  first  equation,  Art.  306, 

2x  :  2y  :  :  4  :  2  ; 
whence  x  =  2y. 

Substituting  this  value  of  x  in  the  second  equation,  we  find 
y=2,  and  x=4. 


8.  .Given  \  ^+^  :  <*-g  :  :  64  :  X  j  to  find  c,  and  y. 

I  a??/=63  j 


By  JLr^.  311,  ic+y  :  x—  y  :  :  8  :  1. 

By  Art.  306,  2x  :  2y  :  :  9  :  7. 

9y 
Hence  x  =  -^-. 

Substituting  this  value  of  x  in  the  second  equation,  we  find 
y=±7,  and  x=±9. 

(x3-y3:(x-y)3::Ql:l[  4 
4.  Given  a  and 


From  the  first  equation,  by  division,  Art.  305, 

Sxy(x—  y)  :  (x—yf  :  :  60  :  1. 
Hence  960  :  (x-yf  :  :  60  :  1, 

or  16  :  (x-y?  ::!:!. 


Therefore 

Hence 
and 

By  addition, 

Hence 

Therefore 
and 


RATIO   AND   PROPORTION. 
X  —  ?/=±4. 


5.  Given 


>.  Given  -j     

(  V^- 


4xy=l2SO. 
+  2xy  +?/2= 


x=+20  or  ±16, 
y=±16  or  ±20. 


x-\ry- 


225 


\  to  find  #  and  y. 
=o  ) 

J.W5. 


=4  or  2. 

=2  or  4. 


—  Va— x=V?/— x  ,        ,, 

"y  J-  to  find  x  and  y. 

y—x+  Va—x  :  Va—x  : :  5  :  2 


Ans. 




O 


7.  Given  x+Vx  :  x—Vx  ::  3Vx+6  :%Vx  to  find  x. 

J.W5.  x=9  or  4. 

8.  What  number  is  that  to  which  if  1,  5,  and  13  be  severally 
added,  the  first  sum  shall  be  to  the  second  as  the  second  to  the 
third?  Ans.  8. 

9.  What  number  is  that  to  which  if  a,  5,  and  c  be  severally 
added,  the  first  sum  shall  be  to  the  second  as  the  second  to  the 

third?  .         l2  —  ac 

Ans.  -  . 

a  —  2b-}-c 

10.  What  two  numbers  are  as  2  to  3,  to  each  of  which  if  4 
be  added,  the  sums  will  be  as  5  to  7  ? 

y    11.  What  two  numbers  are  as  m  to  w,  to  each  of  which  if  a 
'be  added,  the  sums  will  be  as  p  to  q  ? 

Ans.  am(P—9) .  an  (P-V) 
mq  —  np  '     mq  —  np  ' 

12.  What  two  numbers  are  those  whose  difference,  sum,  and 
product  are  as  the  numbers  2,  3,  and  5  respectively  ? 


Ans.  2  and  10. 


K  2 


226  ALGEBRA. 

13.  What  two  numbers  are  those  whose  difference,  sum,  and 
product  are  as  the  numbers  m,  n,  and  p  ? 

IP         A     %P 
Ans.  —  ^—  and  —  ±—  . 

n+m  n  —  m 

14.  Find  two  numbers,  the  greater  of  which  shall  be  to  the 
less  as  their  sum  to  42,  and  as  their  difference  to  6. 

Ans.  32  and  24. 

15.  Find  two  numbers,  the  greater  of  which  shall  be  to  the 
less  as  their  sum  to  a  and  their  difference  to  b. 


2 

16.  There  are  two  numbers  which  are  in  the  ratio  of  3  to  2, 
the  difference  of  whose  fourth  powers  is  to  the  sum  of  their 
cubes  as  26  to  7.     Kequired  the  numbers.        Ans.  6  and  4. 

17.  What  two  numbers  are  in  the  ratio  of  m  to  w,  the  differ 
ence  of  whose  fourth  powers  is  to  the  sum  of  their  cubes  as 

P  to  qf  o       o  * 

*  A       mp    rnr+n9        n  np    m34-wa 

Ans.  -  -  x  —      —  ,  and  —  x  —      —  ,  . 
q      m*  —  n*  q      m*~n* 

18.  Two  circular  metallic  plates,  each  an  inch  thick,  whose 
diameters  are  6  and  8  inches  respectively,  are  melted  and  form 
ed  into  a  single  circular  plate  1  inch  thick.    Find  its  diameter, 
admitting  that  the  area  of  a  circle  varies  as  the  square  of  its 
diameter. 

19.  Find  the  radius  of  a  sphere  whose  volume  is  equal  to 
the  sum  of  the  volumes  of  three  spheres  whose  radii  are  3,  4, 
and  5  inches,  admitting  that  the  volume  of  a  sphere  varies  as 
the  cube  of  its  radius. 

20.  Find  the  radius  of  a  sphere  whose  volume  is  equal  to 
the  sum  of  the  volumes  of  three  spheres  whose  radii  are  r,  r', 
and  r". 


PROGRESSIONS.  227 


CHAPTER  XYI. 

PROGRESSIONS. 
ARITHMETICAL    PROGRESSION. 

320.  An  arithmetical  progression  is  a  series  of  quantities  which 
increase  or  decrease  by  a  common  difference.     Thus  the  fol 
lowing  series  are  in  arithmetical  progression : 

1,3,5,7,9,... 

20,  17,  14,  11,  8,  ... 

a,  a+c?,  a-\-2d,  a  +  3c?,  . . . 

a,  a— d,  a—2d,  a— 3c?,  . . . 

In  the  first  example  the  common  difference  is  -f2,  and  the 
series  forms  an  increasing  arithmetical  progression ;  in  the  sec 
ond  example  the  common  difference  is  —3,  and  the  series  forms 
a  decreasing  arithmetical  progression.  In  the  third  example  the 
common  difference  is  +c?,  and  in  the  fourth  example  it  is  —  d. 

321.  In  an  arithmetical  progression  having  a  finite  number 
of  terms,  there  are  five  quantities  to  be  considered,  viz.,  the  first 
term,  the  last  term,  the  number  of  terms,  the  common  differ 
ence,  and  the  sum  of  the  terms.     When  any  three  of  them  are 
given,  the  other  two  may  be  found.     We  will  denote 

the  first  term  by  a, 

the  last  term  by  Z, 

the  number  of  terms  by  n, 

the  common  difference  by  c?, 

and  the  sum  of  the  terms  by        s. 

The  first  term  and  the  last  term  are  called  the  extremes,  and 
all  the  other  terms  are  called  arithmetical  means. 

322.  In  an  arithmetical  progression  the  last  term  is  equal  to  the 
first  term  plus  the  product  of  the  common  difference  ~by  the  number 
of  terms  less  one. 


228  ALGEBRA. 

Let  the  terms  of  the  series  be  represented  by 
a,  a+d,  a  +  2d,  a  +  8t/,  a  +  4cZ,  etc. 

Since  the  coefficient  of  d  in  the  second  term  is  1,  in  the  third 
term  2,  in  the  fourth  term  3,  and  so  on,  the  nth  term  of  the 
series  will  be 


or  l=a+(n  —  l)c?, 

in  which  d  is  positive  or  negative  according  as  the  series  is  an 

increasing  or  a  decreasing  one. 

323.  The  sum  of  any  number  of  terms  in  arithmetical  pr  ogres* 
sion  is  equal  to  one  half  the  sum  of  the  two  extremes  multiplied  by 
the  number  of  terms. 

The  term  preceding  the  last  will  be  I—  d,  the  term  preceding 
that  I—  2d,  and  so  on.  If  the  terms  of  the  series  be  written  in 
the  reverse  order,  the  sum  will  be  the  same  as  when  written  in 
the  direct  order.  Hence  we  have 


—  &Q+  ---- 
Adding  these  equations  term  by  term,  we  have 


Here  a+l  is  taken  n  times  ;  hence 


n 
or  s—-(a+T). 

£k 

324.  Tn  an  arithmetical  progression  the  sum  of  the  extremes  is 
equal  to  the  sum  of  any  tivo  terms  equidistant  from  the  extremes. 

This  principle  follows  from  the  preceding  demonstration. 
It  may  also  be  shown  independently  as  follows  : 
The  rath  term  from  the  beginning  is  a  +  (m—I)d. 
The  rath,  term  from  the  end  is  I—  (m—  l)d. 

And  the  sum  of  these  terms  is  a-\-l. 

325.  To  insert  any  number  of  arithmetical  means  betiveen  two 
given  terms. 

The  whole  number  of  terms  in  the  series  consists  of  the  two 


PROGRESSIONS. 


229 


extremes  and  all  the  intermediate  terms.  If,  ihen,  m  repre 
sents  the  number  of  means,  m  +  2  will  be  the  whole  number 
of  terms.  • 

Substituting  m-f  2  for  n  in  the  formula,  Art.  322,  we  have 


or 


d=  --  -=the  common  difference. 


whence  the  required  means  are  easily  obtained  by  addition. 
326.  The  two  equations 


contain  five  quantities,  a,  Z,  TI,  d,  5,  of  which  any  three  being 
given,  the  other  two  can  be  found.  We  may  therefore  have 
ten  different  cases,  each  requiring  the  determination  of  two  dif 
ferent  formulae.  These  formulae  are  exhibited  in  the  following 
table,  and  should  be  verified  by  the  student. 


No. 
1. 

2. 
3. 

4. 
5. 

7. 
8. 
9. 
10. 

Given. 

a,  cZ,  ?i, 

?,    4  71, 

a,  Z,  ft, 
a,  71,  5, 

71,  d,  5, 
Z,    71,  5, 

a,  cZ,  Z, 
a,  Z,  5, 
a,  d,  5, 

z,  (U 

Re 
quired. 

a,  5, 

cZ,Z, 
a,  Z, 

<7,  C?, 
71,5, 
71,  CZ, 
Z,   71, 

rf        ^a- 

Formulae. 
^  -1-71  T2Z  (T?  1W1 

?*-!' 

5      0-l)rZ 

5  —  .^^(^-l-Zj. 

7     25 
Z  =  a. 

71 

5      (V?_l")r7 

71               2           ' 

25       _ 

a—         I  • 

»'        2      • 
2»^_2s 

71 

Z-a 

71  —  -            1  .  1  • 

tt  —      1             -1  \' 

n(n  —  1) 
(Z+a)(Z  —  a-hcZ) 

d 

25 

cZ       P~a" 

<=-*d±y2ds+^ 

dY.  n-d     2a±V(2a     d)>  +  8ds 

2d 

2Z+d±y(2l+df     $ds 

a    ^c?±'y/(/4-^c/^)2  —  2o!!s 

n  ±—  i  . 

230  ALGEBKA. 

EXAMPLES. 

1.  The  first  term  of  an  arithmetical  progression  is  2,  and  the 
common  difference  is  4 ;  what  is  the  10th  term  ? 

.4ns.  38. 

2.  The  first  term  is  40,  and  the  common  difference  —3  ;  what 
is  the  10th  term? 

3.  The  first  term  is  1,  and  the  common  difference  f ;  what 
is  the  10th  term  ? 

4.  The  first  term  is  1,  and  the  common  difference  —  J ;  what 
is  the  10th  term  ? 

5.  The  first  term  is  5,  the  common  difference  is  10,  and  the 
number  of  terms  is  60  ;  what  is  their  sum  ? 

Ans.  18000. 

6.  The  first  term  is  116,  the  common  difference  is  —4,  and 
the  number  of  terms  is  25 ;  what  is  their  sum  ? 

7.  The  first  term  is  1,  the  common  difference  is  f ,  and  the 
number  of  terms  is  12  ;  what  is  their  sum  ? 

8.  The  first  term  is  1-f ,  the  common  difference  is  —  f ;  and 
the  number  of  terms  is  10 ;  what  is  their  sum  ? 

9.  Required  the  number  of  terms  of  a  progression  whose  sum 
is  442,  whose  first  term  is  2,  and  common  difference  3. 

Ans.  17. 

10.  Required  the  first  term  of  a  progression  whose  sum  is 
99,  whose  last  term  is  19,  and  common  difference  2. 

11.  The  sum  of  a  progression  is  1455,  the  first  term  5,  and 
the  last  term  92  ;  what  is  the  common  difference  ? 

12.  Required  the  sum  of  101  terms  of  the  series 

1,  3,  5,  7,  9,  etc.  Ans.  10201. 

13.  Find  the  nth  term  of  the  series 

1,  3,  5,  7,  9,  etc.  Ans.  2n-l. 

14.  Find  the  sum  of  n  terms  of  the  series 

1,  3,  5,  7,  9,  etc.  Ans.  n\ 

15.  Find  the  sum  of  n  terms  of  the  series  of  numbers 

1,  2,  3,  4,  5,  etc. 

Ans. 


PROGRESSIONS.  231 

16.  Find  the  sum  of  n  terms  of  the  series 

2,  4,  6,8,  etc.  Ans.  n(n  +  l). 

17.  Find  6  arithmetical  means  between  1  and  50. 

18.  Find  7  arithmetical  means  between  -J-  and  3. 

19.  A  body  falls  16  feet  during  the  first  second,  and  in  each 
succeeding  second  32  feet  more  than  in  the  one  immediately 
preceding ;  if  it  continue  falling  for  20  seconds,  how  many  feet 
will  it  pass  over  in  the  last  second,  and  how  many  in  the  whole 
time? 

Ans.  624  feet  in  the  last  second,  and 
6400  feet  in  the  whole  time. 

20.  One  hundred  stones  being  placed  on  the  ground  in  a 
straight  line  at  the  distance  of  two  yards  from  each  other,  how 
far  will  a  person  travel  who  shall  bring  them  one  by  one  to  a 
basket  which  is  placed  two  yards  from  the  first  stone? 

Ans.  20200  yards. 

PROBLEMS. 

327.  When  of  the  five  quantities  a,  /,  w,  c?,  s,  no  three  are 
directly  given,  it  may  be  necessary  to  represent  the  series  by 
the  use  of  two  unknown  quantities.  The  form  of  the  series 
which  will  be  found  most  convenient  will  depend  upon  the 
conditions  of  the  problem.  If  x  denote  the  first  term  and  y 
the  common  difference,  then 

x,  x+y,  x+Zy,  x+3y,  etc., 
•will  represent  a  series  in  arithmetical  progression. 

It  will,  however,  generally  be  found  most  convenient  to  rep 
resent  the  series  in  such  a  manner  that  the  common  difference 
may  disappear  in  taking  the  sum  of  the  terms.  Thus  a  pro 
gression  of  three  terms  may  be  represented  by 

x-y,  x,  x  +  y; 

one  of  four  terms  by  x— 3y,  %—y,  %+y,  x+Sy; 
one  of  five  terms  by    x  —  2y,  x—y,  x,  x+y,  x  +  2y. 

Prob.  1.  A  number  consisting  of  three  digits  which  are  in 
arithmetical  progression,  being  divided  by  the  sum  of  its  digits, 
gives  a  quotient  26 ;  and  if  198  be  added  to  it,  the  digits  will 
be  inverted  ;  required  the  number.  Ans.  234. 


232  ALGEBRA. 

Prob.  2.  Find  three  numbers  in  arithmetical  progression  the 
sum  of  whose  squares  shall  be  1232,  and  the  square  of  the 
mean  greater  than  the  product  of  the  two  extremes  by  16. 

Ans.  16,  20,  and  24. 

Prob.  3.  Find  three  numbers  in  arithmetical  progression  the 
sum  of  whose  squares  shall  be  a,  and  the  square  of  the  mean 
greater  than  the  product  of  the  two  extremes  by  b. 

/a-'2b       /T        /a-2b 
Ans.  Y — g Vb;  y      ^     ;  and  y 

Prob.  4.  Find  four  numbers  in  arithmetical  progression 
whose  sum  is  28,  and  continued  product  585. 

Ans.  1,  5,  9,  13. 

Prob.  5.  A  sets  out  for  a  certain  place,  and  travels  1  mile 
the  first  day,  2  the  second,  3  the  third,  and  so  on.  In  five  days 
afterward  B  sets  out,  and  travels  12  miles  a  day.  How  long 
will  A  travel  before  he  is  overtaken  by  B  ? 

Ans.  8  or  15  days. 

This  is  another  example  of  an  equation  of  the  second  de 
gree,  in  which  the  two  roots  are  both  positive.  The  following 
diagram  exhibits  the  daily  progress  of  each  traveler.  The  di 
visions  above  the  horizontal  line  represent  the  distances  trav 
eled  each  day  by  A ;  those  below  the  line  the  distances  trav 
eled  by  B. 

A.  12345    6    7     8     9      10     11      12      13       14        15 


UNI 

1     1 

1 

1 

1 

1 

1 

1 

1 

1 

2 

3 

4 

1 
5 

1 
6 

7 

1 
8 

1 
9       1 

B. 

It  is  readily  seen  from  the  figure  that  A  is  in  advance  of  B 
until  the  end  of  his  8th  day,  when  B  overtakes  and  passes  him. 
After  the  12th  day,  A  gains  upon  B,  and  passes  him  on  the 
15th  day,  after  which  he  is  continually  gaining  upon  B,  and 
could  not  be  again  overtaken. 

Prob.  6.  A  goes  1  mile  the  first  day,  2  the  second,  and  so 
on.  B  starts  a  days  later,  and  travels  b  miles  per  day.  How 
long  will  A  travel  before  he  is  overtaken  by  B  ? 


_-- 
Ans.  -    -  '  —  -  days. 


PROGRESSIONS.  233 

In  what  case  would  B  never  overtake  A? 


For  instance,  in  the  preceding  example,  if  B  had  started  one 
day  later,  he  could  never  have  overtaken  A. 

Prob.  7.  A  traveler  set  out  from  a  certain  place  and  went  1 
mile  the  first  day,  3  the  second,  5  the  third,  and  so  on.  After 
he  had  been  gone  three  days,  a  second  traveler  sets  out,  and 
goes  12  miles  the  first  day,  13  the  second,  and  so  on.  After 
how  many  days  will  they  be  together  ? 

Ans.  In  2  or  9  days. 

Let  the  student  illustrate  this  example  by  a  diagram  like  the 
preceding. 

Prob.  8.  A  and  B,  165  miles  distant  from  each  other,  set  out 
with  a  design  to  meet.  A  travels  1  mile  the  first  day,  2  the 
second,  3  the  third,  and  so  on.  B  travels  20  miles  the  first 
day,  18  the  second,  16  the  third,  and  so  on.  In  how  many 
days  will  they  meet?  Ans.  10  or  33  days. 

GEOMETEICAL   PROGRESSION. 

328.  A  geometrical  progression  is  a  series  of  quantities  each  of 
which  is  equal  to  the  product  of  the  preceding  one  by  a  constant 
factor. 

The  constant  factor  is  called  the  ratio  of  the  series. 

329.  When  the  first  term  is  positive,  and  the  ratio  greater 
than  unity,  the  series  forms  an  increasing  geometrical  progres 
sion,  as 

2,  4,  8,  16,  32,  etc., 
in  which  the  ratio  is  2. 

When  the  ratio  is  less  than  unity,  the  series  forms  a  decreas 
ing  geometrical  progression,  as 

81,  27,  9,  3,  etc., 
in  which  the  ratio  is  -J. 

330.  In  a  geometrical  progression  having  a  finite  number 
of  terms,  there  ^XQ  five  quantities  to  be  considered,  viz.,  the  first 


234  ALGEBRA. 

term,  the  last  term,  the  number  of  terms,  the  ratio,  and  the  sum 

of  the  terms.     When  any  three  of  these  are  given,  the  other 

two  may  be  found.     We  will  denote 

the  first  term  by  a, 

the  last  term  by  lt 

the  number  of  terms  by          n, 
the  ratio  by  r, 

and  the  sum  of  the  terms  by  s. 
The  first  term  and  the  last  term  are  called  the  extremes,  and 

all  the  other  terms  are  called  geometrical  means. 

331.  In  a  geometrical  progression,  the  last  term  Is  equal  to  the 
product  of  the  first  term  by  that  power  of  the  ratio  whose  exponent 
is  one  less  than  the  number  of  terms. 

According  to  the  definition,  the  second  term  is  equal  to  the 
first  multiplied  by  r,  that  is,  it  is  equal  to  ar  /  the  third  term 
is  equal  to  the  second  multiplied  by  r,  that  is,  it  is  equal  to 
ar2  ;  the  fourth  term  is  equal  to  the  third  multiplied  by  r,  that 
is,  it  is  equal  to  ar3;  and  so  on.  Hence  the  nth  term  of  the 
series  will  be  equal  to  arn~l  ;  hence  we  shall  have 

l=arn~l. 

332.  To  find  the  sum  of  any  number  of  terms  in  geometrical 
progression,  multiply  the  last  term  by  the  ratio,  subtract  the  first 
term,  and  divide  the  remainder  by  the  ratio  less  one. 

From  the  definition,  we  have 


Multiplying  this  equation  by  r,  we  have 

rs  =  ar-\-ar2-\-  ....     -f  arl~*  +  arn 
Subtracting  the  first  equation  from  the  second,  member  from 
member,  we  have  rs—s=arn—a. 

arn  —  a 

Hence  s  =  -  —  ; 

r  —  1 

or,  substituting  the  value  of  I  already  found,  we  have 

rl—a 


PROGRESSIONS.  235 

If  we  had  subtracted  the  second  equation  from  the  first,  we 
should  have  found 

•  a — rl 

S=T^? 

which  is  the  most  convenient  formula  when  r  is  less  than 
unity,  and  the  series  is,  therefore,  a  decreasing  one. 

333.  To  find  the  sum  of  a  decreasing  geometrical  series 
when  the  number  of  terms  is  infinite,  divide  the  first  term  by 
unity  diminished  by  the  ratio. 

The  sum  of  the  terms  of  a  decreasing  series  may  be  repre 
sented  by  the  formula 

a— rl 

=T~^' 

Now,  in  a  decreasing  series,  each  term  is  less  than  the  pre 
ceding,  and  the  greater  the  number  of  terms,  the  smaller  will 
be  the  last  term  of  the  series.  If  the  number  of  terms  be  in 
finite,  the  last  term  of  the  series  will  be  less  than  any  assigna 
ble  number,  and  rl  may  be  neglected  in  comparison  with  a. 
In  this  case  the  formula  reduces  to 


s= 


l-r 


334.  To  find  any  number  of  geometrical  means  between  two  given 
lerms. 

In  order  to  solve  this  problem,  it  is  necessary  to  know  the 
ratio.  If  m  represent  the  number  of  means,  m  +  2  will  be  the 
whole  number  of  terms.  Hence,  putting  w  +  2  for  n  in  the 
formula,  Art.  331,  we  have 


whence  we  obtain 


r= 


That  is,  to  find  the  ratio,  divide  the  last  term  by  the  first  term, 
and  extract  the  root  which  is  denoted  by  the  number  of  means  plus 
one.  Having  found  the  ratio,  the  required  means  may  be  ob 
tained  by  continued  multiplication. 


236 


ALGEBRA. 


335.  The  two  equations 


5  = 


arn— a 
7^T~' 


contain  five  quantities,  a,  Z,  ??,  r,  5,  of  which  any  three  being 
given,  the  other  two  can  be  found.  We  may  therefore  have 
ten  different  cases,  each  requiring  the  determination  of  two  quan 
tities,  thus  giving  rise  to  twenty  different  formulae.  The  first 
four  of  the  following  cases  are  readily  solved.  The  fifth  and 
sixth  cases  involve  the  solution  of  equations  of  a  higher  degree 
than  the  second.  When  n  is  not  large,  the  value  of  the  un 
known  quantity  can  generally  be  found  by  a  few  trials.  The 
four  remaining  cases,  when  n  is  the  quantity  sought,  involve 
the  solution  of  an  exponential  equation.  See  Art.  416.  These 
different  cases  are  all  exhibited  in  the  following  table  for  con 
venient  reference. 


No. 

Given. 

Re 
quired. 

Formulas. 

1. 

a,  r,  7i, 

u 

7             j                     arn  —  a 

•  r-l  ' 

2. 
3. 

4. 
6. 

7 

a,Z,  n, 
a,  7i,  s, 

a,  5, 

Z                            Zr»-Z 

1                                 n              n 

T>             (  —  1                 *                          <?    

W     '            IE     J_' 

6. 

Z,  71,5, 

a,r, 

a(s  _  a)n-1  —  Z  (5  —  1  )n~  1  ;  (s  —  Z)rn  —  s?<n  "1  =  —  I. 

8. 
9. 
10. 

a,r,Z, 
a,  Z,  5, 
a,  r,  5, 

5,71, 

r,n, 

Z,71, 

a,  n, 

Ir—a                    lo«;.  Z—  log.  a  t  1 

r  —  1                            log.  r 
s  —  a                              log.  Z  —  log.  a          ^  ^ 

r  —        ,  ,               7i  —  i        /         N     i        /       7N  1  i. 
s  —  Z                        log.  (s  —  a)  —  log.  (,9  —  Z) 

r                                      log.  r 

a  —  lr—  (r—  1)5,    TI_    •          •    T5                          1  1 

PROGRESSIONS.  237 

EXAMPLES. 

1.  Find  the  12th  term  of  the  series  1,  3,  9,  27,  etc. 
We  have  1= arn~~L  =  3u  =  177147,  Ans. 

2.  Given  the  first  term  2,  the  ratio  3,  and  the  number  of 
terms  10  ;  to  find  the  last  term.  Ans.  39366. 

3.  Find  the  sum  of  14  terms  of  the  series  1,  2, 4,  8, 16,  etc. 

firfU f* 

5  =  ^_f?  =  2"-l  =  16383,  A^ 
T—L 

4.  Find  the  sum  of  12  terms  of  the  series  1,  3,  9,  27,  etc. 

Ans.  265,720. 

5.  Given  the  first  term  1,  the  last  term  512,  and  the  sum  of 
the  terms  1023 ;  to  find  the  ratio. 

6.  Given  the  last  term  2048,  the  number  of  terms  12,  and 
the  ratio  2  ;  to  find  the  first  term. 

7.  Find  the  sum  of  6  terms  of  the  series  6,  4J,  3f ,  etc. 

Am.  19||f 

8.  Find  the  sum  of  15  terms  of  the  series  8,  4,  2, 1,  etc. 

Anc     1  ^2  04  T 

Ans.  iCM 

9.  Find  three  geometrical  means  between  2  and  162. 

10.  Find  two  geometrical  means  between  4  and  256. 

11.  Find  three  geometrical  means  between  a  and  b. 

Ans. 

12.  Find  the  value  of  1  +  |-+J+-J-+,  etc.,  to  infinity. 

a          1 


13.  Find  the  value  ofl-f-J-f-|--f--g1f+,  etc.,  to  infinity. 

Ans.  f. 

14.  Find  the  value  ofl-j-J-j-^-f-^-f-,  etc.,  to  infinity, 

15.  Find  the  ratio  of  an  infinite  progression  whose  first  term 
is  1,  and  the  sum  of  the  series  f.  Ans.  -J. 

16.  Find  the  first  term  of  an  infinite  progression  whose  ratio 
is  ^V,  and  the  sum  f .  Ans.  %. 

17.  Find  the  first  term  of  an  infinite  progression  of  which 

the  ratio  is  -,  and  the  sum T. 


18.  Find  the  value  of  the  series  3-f  2+|+,  etc.,  to  infinity. 


238  ALGEBRA. 

19.  Find  the  value  of  the  series  f+l+|+,  etc.,  to  infinity. 

20.  A  gentleman,  being  asked  to  dispose  of  his  horse,  said 
he  would  sell  him  on  condition  of  receiving  one  cent  for  the 
first  nail  in  his  shoes,  two  cents  for  the  second,  and  so  on, 
doubling  the  price  of  every  nail  to  32,  the  number  of  nails  in 
his  four  shoes.     What  would  the  horse  cost  at  that  rate  ? 

Ans.  $42,949,672.95. 

PEOBLEMS. 

Prob.  1.  Find  three  numbers  in  geometrical  progression  such 
that  their  sum  shall  be  21,  and  the  sum  of  their  squares  189. 
Denote  the  first  term  by  x  and  the  ratio  by  y  ;  then 
*  +  0^+0^=21,  (1.) 


Transposing  xy  in  Eq.  (1),  squaring,  and  reducing,  we  have 

(3.) 


/* 
Comparing  (2)  and  (3),     xy=§,  or  #=-. 

Substituting  this  value  of  x  in  Eq.  (1),  and  reducing,  we  have 


Whence  y=2  or  J,  and  x=3  or  12. 

The  terms  are  therefore  3,  6,  and  12,  or  12,  6,  and  3. 

Prob.  2.  Find  four  numbers  in  geometrical  progression  such 
that  the  sum  of  the  first  and  second  shall  be  15,  and  the  sum 
of  the  third  and  fourth  60. 

By  the  conditions,  £c+a^  =  15,  (1.) 


Multiplying  Eq.  (1)  by  7/2,  we  have 

ay2  +  ay*  =  150*  =  60. 

Therefore  2/2=4,  and  y=±2. 

Also  cc±2x  =  15; 

therefore  x=5  or  —15. 

Taking  the  first  value  of  x  and  the  corresponding  value  of 
?/,  we  obtain  the  series      5,  10,  20,  and  40. 

Taking  the  second  value  of  x  and  the  corresponding  value 
of  y,  we  obtain  the  series  —15,  +30,  —60,  and  +120; 


PROGRESSIONS.  239 

which  numbers  also  perfectly  satisfy  the  problem  understood 
algebraically.  If,  however,  it  is  required  that  the  terms  of  the 
progression  be  positive,  the  last  value  of  x  would  be  inapplica 
ble  to  the  problem,  though  satisfying  the  algebraic  equation. 

Several  of  the  following  problems  also  have  two  solutions,  if 
we  admit  negative  values. 

Prob.  3.  Find  three  numbers  in  geometrical  progression  such 
that  their  sum  shall  be  210,  and  the  last  shall  exceed  the  first 
by  90.  Ans.  30,  60,  and  120. 

Prob.  4.  Find  three  numbers  in  geometrical  progression  such 
that  their  sum  shall  be  42,  and  the  sum  of  the  first  and  last 
shall  be  34.  Ans.  2,  8,  and  32. 

Prob.  5.  Find  three  numbers  in  geometrical  progression  such 
that  their  continued  product  may  be  64,  and  the  sum  of  their 
cubes  584.  Ans.  2,  4,  and  8. 

Prob.  6.  Find  four  numbers  in  geometrical  progression  such 
that  the  difference  between  the  first  and  second  may  be  4,  and 
the  difference  between  the  third  and  fourth  36. 

Ans.  2,  6, 18,  and  54. 

Prob.  7.  Find  four  numbers  in  geometrical  progression  such 
that  the  sum  of  the  first  and  third  may  be  a,  and  the  sum  of 
the  second  and  fourth  may  be  b. 

a3         tfl         abz  b3 

and 


Prob.  8.  Find  four  numbers  in  geometrical  progression  such 
that  the  fourth  shall  exceed  the  second  by  24,  and  the  sum  of 
the  extremes  shall  be  to  the  sum  of  the  means  as  7  to  3. 

Ans.  1,  3,  9,  and  27. 

Prob.  9.  The  sum  of  $700  was  divided  among  four  persons, 
whose  shares  were  in  geometrical  progression,  and  the  differ 
ence  between  the  greatest  and  least  was  to  the  difference  be 
tween  the  means  as  37  to  12.  What  were  their  respective 
shares  ?  Ans.  108, 144, 192,  and  256. 

^  Prob.  10.  Find  six  numbers  in  geometrical  progression  such 
that  their  sum  shall  be  1365,  and  the  sum  of  the  third  and 
fourth  shall  be  80.  Ans.  1,  4, 16,  64,  256,  and  1024. 


240  ALGEBKA. 


CHAPTER  XVII. 

CONTINUED    FRACTIONS. — PERMUTATIONS  AND   COMBINATIONS, \ 

336.  A  continued  fraction  is  one  whose  numerator  is  unity, 
and  its  denominator  an  integer  plus  a  fraction,  whose  numera 
tor  is  likewise  unity,  and  its  denominator  an  integer  plus  a 
fraction,  and  so  on. 

The  general  form  of  a  continued  fraction  is 


c+1 


d+j  etc. 

When  the  number  of  terms  a,  &,  c,  etc.,  \sfinite,  the  continued 
fraction  is  said  to  be  terminating  ;  such  a  continued  fraction 
may  be  reduced  to  an  ordinary  fraction  by  performing  the  op 
erations  indicated. 

337.  To  convert  any  given  fraction  into  a  continued  fraction. 

Let  -  be  the  given  fraction  ;  divide  m  by  n  ;  let  A  be  the 
n 

quotient,  and  p  the  remainder  :  thus, 


n  n  n 

P 
"Divide  n  by^>;  let  a  be  the  quotient,  and  q  the  remainder: 

thus, 

n  a        .1 

—  =a+--=a-\  —  . 

P          P          P_ 
2 

Similarly,  ^  =  6+-=6+-«, 

r 


CONTINUED  FRACTIONS.  241 

and  so  on,  so  that  we  have 


Z>  +  ,  etc. 

We  see,  then,  that  to  convert  a  given  fraction  into  a  contin 
ued  fraction,  we  proceed  as  if  we  were  finding  the  greatest  com 
mon  divisor  of  the  numerator  and  denominator ;  and  we  must, 
therefore,  at  last  arrive  at  a  point  where  the  remainder  is  zero, 
and  the  operation  terminates ;  hence  every  rational  fraction  can 
"be  converted  into  a  terminating  continued  fraction. 

Ex.  1.  Transform  -       into  a  continued  fraction. 


22  +  1 

Ex.  2.  Transform  |44  into  a  continued  fraction. 

Ans. 


1  +  1 


1  +  1 


Ex.  3.  Transform  |4?  into  a  continued  fraction. 

1 


Ans. 


2  +  1 


_ 

2  +  1 


Ex.  4.  Transform  |4r  into  a  continued  fraction. 
Ex.  5.  Transform  -f-g-J-  into  a  continued  fraction. 
Ex.  6.  Transform  -J|--  into  a  continued  fraction. 


338.  To  find  the  value  of  a  terminating  continued  fraction, 
Ex.  1.  Find  the  value  of  the  continued  fraction 


_ 

s+i- 

Beginning  with  the  last  fraction,  we  have 

L 


242  ALGEBRA. 


Hence 


Therefore 


and 


Ex.  2.  Find  the  value  of  the  continued  fraction 


2  +  1 


3  +  1 


2  +  1 


Ex-  3.  Find  the  value  of  the  continued  fraction 
1 


3  +  1 


2  +  1 


2+i 
Ex.  4.  Find  the  value  of  the  continued  fraction 


l+ 


2  +  1 


1  +  1 


1+1 

1+A- 

339.  To  find  the  value  of  an  infinite  continued  fraction. 

Let  the  fraction  be 

1 


a  +  1 


c+,  etc. 

An  approximate  value  of  this  fraction  is  obtained  by  omit 
ting  all  its  terms  beyond  any  assumed  fraction,  and  obtaining 
the  value  of  the  resulting  fraction,  as  in  the  previous  article. 


CONTINUED   FRACTIONS.  243 

Thus  we  obtain 
1st  approximate  value,  - ; 


2d  approximate  value,  o+l=- 

6 
1_ 
3d  approximate  value,  a+1  ,      1 

7.   ,    -i  OC-\-L 


(bc  +  l)d+b 

4th  approximate  value.  -7-^ — q^— ; — £-= , — r 

'  (ab+I)cd+ad+ab+l,  etc. 

340.  The  fractions  formed  by  taking  one,  two,  three,  etc., 
of  the  quotients  of  the  continued  fraction  are  called  converging 
fractions,  or  conver gents. 

The  convergents,  taken  in  order,  are  alternately  less  and  great 
er  than  the  continued  fraction. 

The  first  convergent  -  is  too  great,  because  the-  denominator 
a  j 

a  is  too  small ;  the  second  convergent  —r — --  is  too  small,  be 
cause  a-f-y  is  too  great,  and  so  on. 

341.  When  a  fraction  has  been  transformed  into  a  continued 
fraction,  its  approximate  value  may  be  found  by  taking  a  few 
of  the  first  terms  of  the  continued  fraction. 

Thus  an  approximate  value  of  -J4r  is  -J,  which  is  the  first 
term  of  its  continued  fraction. 

By  taking  two  terms,  we  obtain  |4,  which  is  a  nearer  ap 
proximation  ;  and  three  terms  would  give  a  still  more  accu 
rate  value. 

Ex.  1.  Find  approximate  values  of  the  fraction  1513/3. 

Ans.  i,  I,  ||. 

Ex.  2.  Find  approximate  values  of  the  fraction 

Ex.  3.  Find  approximate  values  of  the  fraction 


244  ALGEBKA. 

342.  By  the  preceding  method  we  are  enabled  to  discover 
the  approximate  value  of  a  fraction  expressed  in  large  num 
bers,  and  this  principle  has  some  important  applications,  par 
ticularly  in  Astronomy. 

Ex.  4.  The  ratio  of  the  circumference  of  a  circle  to  its  diam- 
ter  is  3.1415926.  Find  approximate  values  for  this  ratio. 


22       333      355 

.    -y, 


Ex.  5.  The  length  of  the  tropical  year  is  365d  5h.  48m.  48s. 
Find  approximate  values  for  the  ratio  of  5  A.  48  ra.  485.  to  24 
hours. 

Ans.  J,  A,  A,  T3TTT- 

Ex.  6.  In  87969  years  the  earth  makes  277287  conjunctions 
with  Mercury.  Find  approximate  values  for  the  ratio  of  87969 
to  277287. 


Ex.  7.  In  57551  years  the  earth  makes  36000  conjunctions 
with  Venus.  Find  approximate  values  for  the  r^,tio  of  57551 
to  36000. 

Ans.  |,  fjf 

Ex.  8.  In  295306  years  the  moon  makes  3652422  synodical 
revolutions.  Find  an  approximate  value  for  the  ratio  of  295306 
to  3652422.  Ans.  -j^. 

Ex.  9.  One  French  metre  is  equal  to  3.2809  English  feet. 
Find  approximate  values  for  the  ratio  of  a  metre  to  a  foot. 

Ex.  10.  One  French  kilogramme  is  equal  to  2.2046  pounds 
avoirdupois.  Find  approximate  values  for  the  ratio  of  a  kilo 
gramme  to  a  pound. 

Ex.  11.  One  French  litre  is  equal  to  0.2201  English  gallons. 
Find  approximate  values  for  the  ratio  of  a  litre  to  a  gallon. 

PERMUTATIONS  AND  COMBINATIONS. 

343.  The  different  orders  in  which  things  can  be  arranged 
are  called  their  permutations.  In  forming  permutations,  all  of 
the  things  or  a  part  only  may  be  taken  at  a  time. 

Thus  the  permutations  of  the  three  letters  a,  6,  c,  taken  all 
together,  are 

abc,  acb.  bac,  bca,  cab,  cba. 


PEKMUTATIONS  AND   COMBINATIONS.  245 

The  permutations  of  the  same  letters  taken  two  at  a  time  are 

ab,  ac,  ba,  be,  ca,  cb. 

The  permutations  of  the  same  letters  taken  one  at  a  time  are 

a,  b,  c. 

344.  The  number  of  permutations  of  n  things  taken  m  at  a  time 
is  equal  to  the  continued  product  of  the  natural  series  of  numbers 
from  n  down  to  n—m  +  l. 

Suppose  the  things  to  be  n  letters,  a,  5,  c,  d 

The  number  of  permutations  of  n  letters,  taken  singly  or 
one  at  a  time,  is  evidently  equal  to  the  number  of  letters,  or 
to  n. 

If  we  wish  to  form  all  the  permutations  of  n  letters  taken 
two  at  a  time,  we  must  write  after  each  letter  each  of  the 
Ti—1  remaining  letters.  We  shall  thus  obtain  n(n—l)  permu 
tations. 

If  we  wish  to  form  all  the  permutations  of  n  letters  taken 
three  at  a  time,  we  must  write  after  each  of  the  permutations 
of  n  letters  taken  two  at  a  time  each  of  the  n— 2  remaining  let 
ters.  We  shall  thus  obtain  n(n—  l)(n— 2)  permutations. 

In  the  same  manner  we  shall  find  that  the  number  of  permu 
tations  of  n  letters  taken  four  at  a  time  is 
n(n-l)(n-2)(n-3). 

Hence  we  may  conclude  that  the  number  of  permutations 
of  n  letters  taken  m  at  a  time  is 

n(n-l)(n-2)(n—S) (n-m  +  1). 

345.  The  number  of  permutations  of  n  things  taken  all  together 
is  equal  to  the  continued  product  of  the  natural  series  of  numbers 
from  1  to  n. 

If  we  suppose  that  each  permutation  comprehends  all  the  n 
letters;  that  is,  if  m  —  n,  the  preceding  formula  becomes 

n(n-l)(n-2) 3x2x1; 

or,  inverting  the  order  of  the  factors, 

1.2.3.4.  .  .  .     (7i-l)w, 

which  expresses  the  number  of  permutations  of  n  things  taken 
all  together. 


246  ALGEBRA. 

For  the  sake  of  brevity,  1.2.3.4....  (n— l)n  is  often  de- 
noted  by  \n;  that  is,  \n  denotes  the  product  of  the  natural  num 
bers  from  1  to  n  inclusive. 

346.  The  combinations  of  things  are  the  different  collections 
which  can  be  formed  out  of  them  without  regarding  the  order 
in  which  the  things  are  placed. 

Thus  the  three  letters  a,  6,  c,  taken  all  together,  form  but  one 
combination,  abc. 

Taken  two  and  two,  they  form  three  combinations,  ab,  ac,  be. 

347.  The  number  of  combinations  of  n  things,  taken  m  at  a  time, 
is  equal  to  the  continued  product  of  the  natural  series  of  numbers 
from  n  down  to  n  —  m-\-\  divided  by  the  continued  product  of  the 
natural  series  of  numbers  from  1  to  m. 

The  number  of  combinations  of  n  letters  taken  separately, 
or  one  at  a  time,  is  evidently  n. 

The  number  of  combinations  of  n  letters  taken  two  at  a 

.    n(n  —  l) 
time  is  •  '. 

L  ,& 

For  the  number  of  permutations  of  n  letters  taken  two  at  a 
time  is  n(n — 1),  and  there  are  two  permutations  (ab,  bd)  corre 
sponding  to  one  combination  of  two  letters ;  therefore  the  num 
ber  of  combinations  will  be  found  by  dividing  the  number  of 
permutations  by  2. 

The  number  of  combinations  of  n  letters  taken  three  at  a 

.    n(n-l)(n-2) 
time  is  -— - —     yv '-. 

For  the  number  of  permutations  of  n  letters  taken  three  at 
a  time  is  n(n  —  ~L)(n  —  2),  and  there  are  1.2.3  permutations  for 
one  combination  of  these  letters ;  therefore  the  number  of  com 
binations  will  be  found  by  dividing  the  number  of  permuta 
tions  by  1.2.3. 

In  the  same  manner  we  shall  find  the  number  of  combina 
tions  of  n  letters  taken  m  at  a  time  to  be 

77(71— l)(n  — 2)  .  .  .  .     (n  —  m- 
1.2.3  .  m 


PERMUTATIONS   AND   COMBINATIONS.  247 

EXAMPLES. 

1.  How  many  different  permutations  may  be  formed  of  8 
letters  taken  5  at  a  time  ?  Arts.  8.7.6.5.4= 6720. 

2.  How  many  different  permutations  may  be  formed  of  the 
26  letters  of  the  alphabet  taken  4  at  a  time  ? 

Ans.  358800. 

3.  How  many  different  permutations  may  be  formed  of  12 
letters  taken  6  at  a  time  ?  Ans.  665280. 

4.  How  many  different  permutations  may  be  formed  of  8 
things  taken  all  together? 

Ans.  1.2.3.4.5.6.7.8=40320. 

5.  How  many  different  permutations  may  be  made  of  the 
letters  in  the  word  Roma  taken  all  together? 

6.  How  many  different  permutations  may  be  made  of  the 
letters  in  the  word  virtue  taken  all  together? 

7.  What  is  the  number  of  different  arrangements  which  can 
be  formed  of  12  persons  at  a  dinner- table  ? 

Ans.  479001600. 

8.  How  many  different  combinations  may  be  formed  of  6 
letters  taken  3  at  a  time  ? 

6.5.4     on 

Ans-  ro=2a 

9.  How  many  different  combinations  may  be  formed  of  8 
letters  taken  4  at  a  time  ?  Ans.  70. 

10.  How  many  different  combinations  may  be  formed  of  10 
letters  taken  6  at  a  time?  Ans.  210. 

11.  A  telegraph  has  m  arms,  and  each  arm  is  capable  of  n 
distinct  positions;  find  the  total  number  of  signals  which  can 
be  made  with  the  telegraph. 

12.  How  many  different  numbers  can  be  formed  with  the 
digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  each  of  these  digits  occurring  once, 
and  only  once,  in  each  number  ? 


248  ALGEBKA. 


CHAPTER  XVIII. 

BINOMIAL    THEOREM. 

348.  The  binomial  theorem,  or  binomial  formula,  is  a  formula 
discovered  by  Newton,  by  means  of  which  we  may  obtain  any 
power  of  a  binomial  x+a,  without  obtaining  the  preceding 
powers. 

349.  By  actual  multiplication,  we  find  the  successive  powers 
of  x+a  to  be  as  follows: 

(x  +  a)2  —  x2  -f  2ax  +  a2, 
(x  +  a)3  =  x3  +  3aa?2  +  3a2#  +  a3, 
(x  +  a)4  =  x4  +  4ax3  +  6a2x2  +  4a3x  +  a4, 
(SB  +  a)5  =  ^5  +  5acc4  +  lOaV  +  10«3x2  +  5a*x  +  a6. 
The  powers  of  a?  —  a,  found  in  the  same  manner,  are  as  fol 
lows: 


On  comparing  the  powers  of  x-\-a  with  those  of  x—  a,  we 
perceive  that  they  only  differ  in  the  signs  of  certain  terms.  In 
the  powers  of  x+a,  all  the  terms  are  positive.  In  the  powrers 
of  x  —  a,  the  terms  containing  the  odd  powers  of  a  have  the  sign 
minus,  while  the  terms  containing  the  even  powers  have  the 
sign  plus.  The  reason  of  this  is  obvious;  for,  since  —a  is  the 
only  negative  term  of  the  root,  the  terms  of  the  power  can  only 
be  rendered  negative  by  a.  A  term  which  contains  the  factor 
—a  an  even  number  of  times  will  therefore  be  positive;  if  it 
contain  it  an  odd  number  of  times  it  must  be  negative.  Hence 
it  appears  that  it  is  only  necessary  to  seek  for  a  method  of  ob 
taining  the  powers  of  x-\-a,  for  these  will  become  the  powers 
of  x—  a  by  simply  changing  the  signs  of  the  alternate  terms. 


BINOMIAL   THEOREM. 


249 


350.  Law  of  the  Exponents. — The  exponents  of  x  and  of  a  in 
the  different  powers  follow  a  simple  law.     In  the  first  term  of 
each  power,  x  is  raised  to  the  required  power  of  the  binomial ; 
and  in  the  following  terms  the  exponents  of  x  continually  de 
crease  by  unity  to  zero,  while  the  exponents  of  a  increase  by 
unity  from  zero  up  to  the  required  power  of  the  binomial. 

351.  Law  of  the  Coefficients. — The  coefficient  of  the  first  term 
is  unity ;  that  of  the  second  term  is  the  exponent  of  the  power ; 
and  the  coefficients  of  terms  equidistant  from  the  extremes  are 
equal  to  each  other;  but  after  the  first  two  terms  it  is  not  ob 
vious  how  to  obtain  the  coefficients  of  the  fourth  and  higher 
powers. 

In  order  to  discover  the  law  of  the  coefficients,  we  will  form 
the  product  of  several  binomial  factors  whose  second  terms  are 
all  different;  thus, 


(x+a)(x+b)=x*+a   x+ab. 


(x+a)(x+b)(x+c)(x+d)=x*+a 


+  ad 
+bd 
+  cd 


x  -\-dbc. 


x2-{-abc 
+  abd 


-\-bcd 


x -\-dbcd. 


In  each  of  these  products  the  exponent  of  x  in  the  first  term 
is  equal  to  the  number  of  binomial  factors,  and  in  the  follow- 
ng  terms  continually  decreases  by  one.  The  coefficient  of  the 
first  term  is  unity ;  the  coefficient  of  the  second  term  is  the  sum  of 
the  second  terms  of  the  binomial  factors  ;  the  coefficient  of  the  third 
term  is  the  sum  of  all  their  products  taken  two  and  two,  and  so  on. 
The  last  term  is  the  product  of  the  second  terms  of  the  binomial 
factors. 


250  ALGEBRA. 

352  We  will  now  prove  that  if  the  laws  of  formation  jusi 
stated  are  true  for  any  power,  they  will  also  hold  true  for  the  forma 
tion  of  the  next  higher  power. 

Suppose  that  we  have  found  the  product  of  ra  binomials 
a?  +  a,  a?  +5,  ....  x+L  Let  Pj  denote  the  sum  of  the  second 
terms  of  the  binomials,  P2  the  sum  of  the  different  products  of 
these  second  terms  taken  two  and  two,  P3  the  sum  of  their 
products  taken  three  and  three,  and  so  on  ;  and  let  Pm  denote 
the  product  of  all  these  second  terms.  The  product  of  the 
given  binomials  will  then  be 


Multiplying  this  polynomial  by  a  new  binomial,  a?+?,  we  ob 
tain  the  following  product  : 


-{-I 


— 1 


+  /P, 


+P1 


l-l 


The  law  of  the  exponents  of  x  remains  the  same.  The  co 
efficient  of  the  first  term  is  still  equal  to  unity,  and  that  of  the 
second  term  is  the  sum  of  the  second  terms  of  the  ra  +  1  bino 
mials.  The  coefficient  of  the  third  term  consists  of  the  sum 
of  the  products  of  the  second  terms  of  the  m  binomial  factors 
taken  two  and  two,  increased  by  the  sum  of  the  same  second 
terms  multiplied  by  Z,  which  is  equivalent  to  the  sum  of  the 
products  of  the  second  terms  of  the  ra+1  binomials  taken  two 
and  two.  The  coefficient  of  the  fourth  term  consists  of  the  sum 
of  the  products  of  the  second  terms  of  the  m  factors  of  the  first 
product  taken  three  and  three,  increased  by  the  sum  of  the 
products  of  their  second  terms  taken  two  and  two  multiplied 
by  7,  which  is  equivalent  to  the  sum  of  the  products  of  the  sec 
ond  terms  of  the  ra  +  1  binomials  taken  three  and  three,  and 
so  on.  The  last  term  is  equal  to  the  product  of  the  second 
terms  of  the  m  binomial  factors  multiplied  by  ?,  which  is  equiv 
alent  to  the  product  of  the  second  terms  of  the  ra  +  1  binomials. 

Hence  the  law  which  was  supposed  true  for  m  factors  is  true 
for  ra  +  1  factors;  and  therefore,  since  it  has  been  verified  for 
two  factors,  it  is  true  for  three ;  being  true  for  three  factors,  it 
is  also  true  for  four,  and  so  on ;  therefore  the  law  is  general. 


BINOMIAL   THEOKEM.  251 

353.  Powers  of  a  Binomial.  —  If  now,  in  the  preceding  bino 
mial  factors,  we  suppose  the  second  terms  to  be  all  equal  to  a, 
the  product  of  these  binomials  will  become  the  rath  power  of 


The  coefficient  of  the  second  term  of  the  product  becomes 
equal  to  a  multiplied  by  the  number  of  factors;  that  is,  it  is 
equal  to  ma. 

The  coefficient  of  the  third  term  reduces  to  a2  repeated  as 
many  times  as  there  are  different  combinations  of  m  letters 

taken  two  and  two  ;  that  is,  to  —  ^—  —  -  a2. 

1.  Z 

The  coefficient  of  the  fourth  term  reduces  to  a3  repeated  as 
many  times  as  there  are  different  combinations  of  m  letters 

.       .    m(m  —  l)(m  —  2)   _ 
taken  three  and  three;  that  is  --  TOO  -  a  »  anc*  so  on- 

L  .  Zt.O 

The  last  term  will  be  am. 

Hence  the  mth  power  of  oc+a  may  be  expressed  as  follows: 

(x  4-  a)m  =m  m~l     m~   ) 


1.2.3 

354.  We  perceive  that  if  the.  coefficient  of  any  terra  be  multi 
plied  by  the  exponent  of  x  in  that  term,  and  the  product  be  divided 
by -the  exponent  of  a  in  that  term  increased  by  unity,  it  will  give  the 
coefficient  of  the  succeeding  term. 

Forming  thus  the  seventh  power  of  x+a,  we  obtain 


We  have  thus  deduced 

Sir  Isaac  Newton's  Binomial  Theorem. 

355.  In  any  power  of  a  binomial  x-\-a,  the  exponent  of  x  begins 
in  the  first  term  with  the  exponent  of  the  power,  and  in  the  follow 
ing  terms  continually  decreases  by  one.     The  exponent  of  a  com 
mences  with  one  in  the  second  term  of  the  power,  and  continually 
increases  by  one. 

The  coefficient  of  the  first  term  is  one,  that  of  the  second  is  the  ex 


252  ALGEBKA. 

ponent  of  the  power  ;  and  if  the  coefficient  of  any  term  be  multi 
plied  by  the  exponent  of  x  in  that  term,  and  divided  by  the  exponent 
of  a  increased  by  one,  it  will  give  the  coefficient  of  the  succeeding  term. 

356.  The  coefficient  of  the  nth  term  from  the  beginning  is  equal 
to  the  coefficient  of  the  nth  term  from  the  end. 

If  we  change  the  places  of  x  and  a,  we  shall  have,  by  the  law 
of  formation, 


(a  +  x)™  =  a™  +  mxa™~^  +        "-     xiam-<i  + 

•L.Z 

m(m-l)(m-2) 

1.2.3 

The  second  member  of  this  equation  is  the  same  as  the  sec 
ond  member  of  the  equation  in  Art.  353,  but  taken  in  a  reverse 
order.  Comparing  the  two,  we  see  that  the  coefficient  of  the 
second  term  from  the  beginning  is  equal  to  the  coefficient  of 
the  second  term  from  the  end  ;  the  coefficient  of  the  third  from 
the  beginning  is  equal  to  that  of  the  third  from  the  end,  and 
so  on.  Hence,  in  forming  any  power  of  a  binomial,  it  is  only 
necessary  to  compute  the  coefficients  for  half  the  terms  ;  we 
then  repeat  the  same  numbers  in  a  reverse  order. 

357.  The  mth  power  ofx-\-a  contains  m  +  ~L  terms.     This  ap 
pears  from  the  law  of  formation  of  the  powers  of  a  binomial 
developed  in  Art.  352.     Thus  the  fourth  power  of  x-}-a  con 
tains  five  terms  ;  the  sixth  power  contains  seven  terms,  etc.' 

358.  The  sum  of  the  coefficients  of  the  terms  in  the  nth  power  of 
x-\-a  is  equal  to  the  nth  power  of  2. 

For,  suppose  cc=l  and  a=l,  then  each  term  of  the  formula 
without  the  coefficients  reduces  to  unity,  and  the  sum  of  the 
terms  is  simply  the  sum   of  the   coefficients.     In  this  case 
(x-\-a)m  becomes  (l  +  l)m,  or  2m. 
Thus  the  coefficients  of  the 

second  power  are  1+2  +  1=  4=22, 

third  "  1  +  3  +  3  +  1=  8=23, 

fourth          "         l+4+6+4+l  =  16  =  24,  etc. 


BINOMIAL   THEOREM.  253 

359.  To  obtain  the  development  of  (x  —  a)m,  it  is  sufficient  to 
change  +a  into  —a  in  the  development  of  (x  +  d)m.  In  con 
sequence  of  this  substitution,  the  terms  which  contain  the  odd 
powers  of  a  will  have  the  minus  sign,  while  the  signs  of  the  re 
maining  terms  will  be  unchanged.  We  shall  therefore  have 


1.2.3 


EXAMPLES. 

1.  Find  the  sixth  power  of  a  -}-b. 
The  terms  without  the  coefficients  are 

a6,  a56,  a462,  a363,  a2£4,  ab5,  b6. 
The  coefficients  are 

1    «  6x5  15x4  20x3  15x2  6x1 
'    '     2    '      3     '  ~T~'  ~T~'~1T 
that  is,         1,  6,     15,       20,        15,         6,          1. 
Prefixing  the  coefficients,  we  obtain 


2.  Find  the  ninth  power  of  a—  b. 
The  terms  without  the  coefficients  are 


,        , 
The  coefficients  are 

9x8  36x7  84x6  126x5  126x4  84x3  36x2  9x1 
2    '      3     '     4     '      5      '       6      '  ~T~'  ~~8~'  ~9~ 
that  is, 

1,  9,   36,      84,       126,       126,        84,        36,         9,         1. 
Prefixing  the  coefficients,  we  obtain 


It  should  be  remembered  that  it  is  only  necessary  to  com 
pute  the  coefficients  of  half  the  terms  independently. 

3.  Find  the  seventh  power  of  a  —  x. 

4.  Find  the  third  term  of  (a  +  b)15. 

5.  Find  the  forty-ninth  term  of  (a—  a;)50. 

6.  Find  the  middle  term  of  (a  -fa;)10. 


254:  ALGEBRA. 

360.  A  Binomial  with  Coefficients.  —  If  the  terms  of  the  given 
binomial  have  coefficients  or  exponents,  we  ma;y  obtain  any 
power  of  it  by  means  of  the  binomial  formula.  For  this  pur 
pose,  each  term  must  be  raised  to  its  proper  power  denoted  by 
the  exponents  in  the  binomial  formula. 

7.  Find  the  fourth  power  of  2x+3a. 

For  convenience,  let  us  substitute  y  for  2x  and  b  for  3a. 
Then  (y  +  fe)4  =y  +  ty*b  +  6y2b2  +  tyb3  4-  fe4. 

Eestoring  the  values  of?/  and  fe, 
the  first  term  will  be  (2ce)4=16£e*, 

the  second  term  will  be     4(2x)3x3a  = 
the  third  term  will  be     6(2x)2  x  (3«)2 
the  fourth  term  will  be    4(2x)x(3a)3  = 
the  fifth  term  will  be  (3a)*  =  81a*. 

Therefore  (2x+3a)4=:16x4  +  96x3a+216x2tt2 

It  is  recommended  to  write  the  three  factors  of  each  term  in 
a  vertical  column,  and  then  perform  the  multiplication  as  indi 
cated  below  : 

Coefficients,          1+4+6       +4+1 

Powers  of  2x,     16x4  +   8x3  +     4x2    +     2x    +   1 

Powers  of  3«,       1     +   3a    +     9a2    +   27  a3 


8.  Find  the  fifth  power  of  2ax—  3b. 

Coefficients,  1        +5          +10  +10  +5        +1 

Powers  of  2aar,      32a5.r5+   16a4a:4    +     Sa5x3     +       4a2ar2     +     2nx    +      1 
Powers  of  -36,      1  36        +     962  27b3        +  81  &*     -2436s 

(2ax  -  36)5  =32a!'x6  -  240a4x46  +  720a3a:362  -  1080a2a:263  +  8  Waxb*  -  24366. 

9,  Find  the  fourth  power  of  2x+5a2. 

Ans.  1  6x4  +  1  60x3a2  +  600x2a4  +  lOOOxa6  +  625a8. 

10.  Find  the  fourth  power  of  x3  +  4?/2. 

11,  Find  the  sixth  power  of  a3  +  Safe. 


12.  Find  the  seventh  power  of  2a  —  Sfe. 

Ans.  128a7  - 


13.  Find  the  fifth  power  of  5a2  — 


BINOMIAL   THEOREM.  255 

14.  Find  the  sixth  power  of  a2x  +  by2. 

15.  Find  the  fifth  power  of  ax—1. 

16.  Find  the  fifth  term  of  (a2-62)12. 

17.  Find  the  fifth  term  of  (Sx^-ty^)9. 

18.  Find  the  sixth  power  of  5  —  ^. 

Powers  and  Roots  of  Polynomials. 

361-  If  it  is  required  to  raise  &  polynomial  to  any  power,  we 
iiiay,  by  substituting  other  letters,  reduce  it  to  the  form  of  a 
binomial.  We  obtain  the  power  of  this  binomial  by  the  gen 
eral  formula  ;  then,  restoring  the  original  letters,  and  perform 
ing  the  operations  indicated,  we  obtain  the  required  power  of 
the  proposed  polynomial. 

Ex.  1.  Let  it  be  required  to  raise  a  +  b-\-c  to  the  third  power. 

If  we  put  b-\-c=m,  we  shall  have 

(a  +  1  4-  c)3  =  (a  4-  m)3  =  a3  +  3a2m  +  3am2  +  m3, 
or  a3  +  Sa2(b  +  c)  +  3a(6  +  c)2  +  (ft  +  c)3. 

Developing  the  powers  of  the  binomial  b-\-c,  and  performing 
the  operations  indicated,  we  obtain 


Ex.  2.  Find  the  fifth  power  of 

Ex.  3.  Find  the  fourth  power  of  a2  — 

Ans.  a8  -  4«7£  +  1  Oa6£2  -  1  6a5ft3  +  1  9a4&4 

-  16a3//  +  10a2^6-4(/&7  +  1*. 
Ex.  4.  Find  the  fifth  power  of  1+2^-f  3x2. 
Ex.  5.  Find  the  sixth  power  of  a 
Ans.  a6-f  6a5&  +  6a5c  + 

60a36c2  +  20a3c3 

+  15a2c*  +  6«Z>5 
4  4-  6ac5  -f  Z>6  4-  Qb 


362.  The  binomial  theorem  will  inform  us  how  to  extract 
any  root  of  a  polynomial.     We  know  that  the  mih  power  of 


256  ALGEBRA. 


x  +  a  is  xm+  max™-^  +  other  terms.  The  first  term  of  the  root 
is,  therefore,  the  mth  root  of  the  first  term  of  the  polynomial. 
Also  the  second  term  of  the  root  may  be  found  by  dividing  the 
second  term  of  the  polynomial  by  mxm~l  ;  that  is,  the  first  term 
of  the  root  raised  to  the  next  inferior  power,  and  multiplied 
by  the  exponent  of  the  given  power.  Hence,  for  extracting 
any  root  of  a  polynomial,  we  have  the  following 

RULE. 

Arrange  the  terms  according  to  the  powers  of  one  of  the  letters, 
and  take  the  mth  root  of  the  first  term  for  the  first  term  of  the  re 
quired  root. 

Subtract  the  mth  power  of  this  term  of  the  root  from  the  given 
polynomial,  and  divide  the  first  term  of  the  remainder  ly  m  times 
the  (m  —  1)  power  of  this  root  ;  the  quotient  will  be  the  second  term 
of  the  root. 

Subtract  the  mth  power  of  the  terms  already  found  from  the  given 
polynomial,  and,  using  the  same  divisor,  proceed  in  like  manner  to 
find  the  remaining  terms  of  the  root. 

Ex.  1.  Find  the  fourth  root  of 


32a3)         — 


Here  we  take  the  fourth  root  of  16a4,  which  is  2a,  for  the 
first  term  of  the  required  root,  subtract  its  fourth  power,  and 
bring  down  the  first  term  of  the  remainder,  —  96a3£.  For  a 
divisor,  we  raise  the  first  term  of  the  root  to  the  third  power 
and  multiply  it  by  4,  making  32a3.  Dividing,  we  obtain  —  Sx 
for  the  second  term  of  the  root.  The  quantity  2a  —  3x,  being 
raised  to  the  fourth  power,  is  found  to  be  equal  to  the  proposed 
polynomial. 

Ex.  2.  Find  the  fifth  root  of 

-40a2f  lOa-1. 

Ans.  2x—~L 


BINOMIAL  THEOREM.  257 

Ex.  3.  Find  the  fourth  root  of 

336o;5+81z8-216x7--56x4+16-224x3  +  64x. 

Am.  3x2— 2z— 2. 

363.  To  extract  any  Root  of  a  Number. — The  preceding  meth 
od  may  be  applied  to  the  extraction  of  any  root  of  a  number. 
Let  n  be  the  index  of  the  root,  n  being  any  whole  number. 
For  a  reason  similar  to  that  given  for  the  square  and  cube  roots, 
we  must  first  divide  the  number  into  periods  of  n  figures  each, 
beginning  at  the  right.  The  left-hand  period  may  contain  less 
than  n  figures.  Then  the  first  figure  of  the  required  root  will 
be  the  nth  root  of  the  greatest  nth  power  contained  in  the  first 
period  on  the  left.  If  we  subtract  the  nth  power  of  this  root 
from  the  given  number,  and  divide  the  remainder  by  n  times 
the  (n— l)th  power  of  the  first  figure,  regarding  its  local  value, 
the  quotient  will  be  the  second  figure  of  the  root,  or  possibly 
a  figure  too  large.  The  result  may  be  tested  by  raising  the 
whole  root  now  found  to  the  nth  power ;  and  if  there  are  other 
figures  they  may  be  found  in  the  same  manner. 

In  the  extraction  of  the  nth  root  of  an  integer,  if  there  is 
still  a  remainder  after  we  have  obtained  the  units'  figure  of  the 
root,  it  indicates  that  the  proposed  number  has  not  an  exact  nth 
root.  We  may,  if  we  please,  proceed  with  the  approximation 
to  any  desired  extent  by  annexing  any  number  of  periods  of 
n  ciphers  each,  and  continuing  the  operation.  We  thus  ob 
tain  a  decimal  part  to  be  added  to  the  integral  part  already 
found. 

So,  also,  if  a  decimal  number  has  no  exact  nth  root,  we  may 
annex  ciphers,  and  proceed  with  the  approximation  to  any  de 
sired  extent,  dividing  the  number  into  periods  commencing 
with  the  decimal  point. 

Ex,  1.  Find  the  fifth  root  of  33554432. 

335.54432(32 
243 

6.3* =405)   925 
325=         33554432. 

Ex.  2.  Find  the  fifth  root  of  4984209207. 


258  ALGEBRA. 

Ex.  3.  Find  the  fifth  root  of  10. 

Ex.  4.  Find  the  fifth  root  off.  Am.  .922. 

364.  When  the  index  of  the  required  root  is  composed  of 
two  factors,  we  may  obtain  the  root  required  by  the  successive 
extraction  of  simpler  roots,  Art.  217.  For  the  mnih  root  of 
any  number  is  equal  to  the  mth  root  of  the  nth  root  of  that 
number. 

Thus  we  may  obtain  the  fourth  root  by  extracting  the  square 
root  of  the  square  root. 

We  may  obtain  the  sixth  root  by  extracting  the  cube  root 
of  the  square  root,  or  the  square  root  of  the  cube  root.  It  is, 
however,  best  to  extract  the  roots  of  the  lowest  degrees  first, 
because  the  operation  is  less  laborious. 

We  may  obtain  the  eighth  root  by  extracting  the  square  root 
three  times  successively.  We  may  obtain  the  ninth  root  by  ex 
tracting  the  cube  root  twice  successively. 

Ex.  1.  Find  the  fourth  root  of 

6a262  +  a4  -  4a36  -  4a  £3  +  64. 

Ex.  2.  Find  the  sixth  root  of 


Ex.  3.  Find  the  eighth  root  of 


SERIES.  259 


CHAPTER  XIX. 

SERIES. 

365.  A  series  is  a  succession  of  terms  each  of  which  is  de° 
rived  from  one  or  more  of  the  preceding  ones  by  a  fixed  law. 
This  law  is  called  the  law  of  the  series.     The  number  of  terms 
of  the  series  is  generally  unlimited.     Arithmetical  and  geomet 
rical  progressions  afford  examples  of  series. 

366.  A  converging  series  is  one  in  which  the  sum  of  the  first 
n  terms  can  not  numerically  exceed  some  finite  quantity,  how 
ever  great  n  may  be. 

Thus,  1,  J,  i,  -J,  TV?  etc-j  is  a  converging  series. 

367.  A  diverging  series  is  one  in  which  n  can  be  taken  so 
large  that  the  sum  of  the  first  n  terms  is  numerically  greater 
than  any  finite  quantity. 

Thus,  1,  2,  3,  4,  5,  6,  etc.,  is  a  diverging  series. 

368.  When  a  certain  number  of  terms  are  given,  and  the 
law  of  the  series  is  known,  we  may  find  any  term  of  the  series, 
or  the  sum  of  any  number  of  terms.     This  may  generally  be 
done  by  the  method  of  differences. 

369.  To  find  the  several  orders  of  differences  for  arty  series: 
Subtract  the  first  term  from  the  second,  the  second  from  the 

third,  the  third  from  the  fourth,  etc. ;  we  shall  thus  form  a  new 
series,  which  is  called  the  first  order  of  differences. 

Subtract  the  first  term  of  this  new  series  from  the  second,  the 
second  from  the  third,  etc. ;  we  shall  thus  form  a  third  series, 
called  the  second  order  of  differences. 

Proceed  in  like  manner  for  the  third,  fourth,  etc.,  orders  of 
differences,  and  so  on  till  they  terminate,  or  are  carried  as  far 
as  may  be  thought  necessary. 


260 


ALGEBRA. 


Ex.  1.  Find  the  several  orders  of  differences  of  trie  series  oi 
square 'numbers  1,  4,  9, 16,  etc. 

Squares. 
1 

4 
9 

16 
25 

Ex.  2.  Find  the  several  orders  of  differences  of  the  series  of 
cube  numbers  1,  8,  27,  etc. 


1st  Diff. 

2d  Diff. 

3dD 

3 

5 

2 

0 

7 

2 

0 

9 

2 

0 

2 

Cubes. 

1st  Diff. 

2d  Diff. 

3d  Diff. 

4th  Diff. 

1 

8 

7 

12 

27 

19 

18 

6 

0 

64 

37 

24 

6 

0 

125 

61 

30 

6 

0 

216 

91 

36 

6 

0 

Ex.  3.  Find  the  several  orders  of  differences  of  the  series  of 
fourth  powers  1,  16,  81,  256,  625,  1296,  etc. 

Ex.  4.  Find  the  several  orders  of  differences  of  the  series  of 
fifth  powers  1,  32,  243,  1024,  3125,  7776, 16807,  etc. 

Ex.  5.  Find  the  several  orders  of  differences  of  the  series  of 
numbers  1,  3,  6,  10,  15,  21,  etc. 

370.   To  find  the  nth  term  of  any  series: 

Let  a,  6,  c,  d,  e,  etc.,  represent  the  proposed  series.  If  we 
subtract  each  term  from  the  next  succeeding  one,  we  shall  ob 
tain  the  first  order  of  differences ;  if  we  subtract  each  term  of 
this  new  series  from  the  succeeding  term,  we  shall  obtain  the 
second  order  of  differences,  and  so  on,  as  exhibited  in  the  fol 
lowing  table : 

4th  Order  of  Differences. 


e_4c?-j-6c— 


Series. 

1st  Diff. 

2d  Differences. 

3d  Order  of  Differences. 

a 

7 

b-a 

0 

c 
d 

c-b 
d-c 

d-2c+b 
e—2d+c 

d—Bc+Sb—  a 
e—3d+Sc—b 

e  —d 

e 

SERIES.  261 

Let  D',  D",  D'",  D"",  etc.,  represent  the  first  terms  of  the 
several  orders  of  differences.     Then  we  shall  have 
T>'—b—  a,  whence  b=a+~D'. 

D"=c-26+a,  "      c 


etc.,  etc. 

The  coefficients  of  the  value  of  c,  the  third  term  of  the  pro 
posed  series,  are  1,  2,  1,  which  are  the  coefficients  of  the  second 
power  of  a  binomial  ;  the  coefficients  of  the  value  of  d,  the 
fourth  term,  are  1,  3,  3,  1,  which  are  the  coefficients  of  the  third 
power  of  a  binomial,  and  so  on.  Hence  we  infer  that  the  co 
efficients  of  the  nth  term  of  the  series  are  the  coefficients  of  the 
(tt—  l)th  power  of  a  binomial.  If  we  denote  the  nth  term  of 
the  series  by  Tn,  we  shall  have 


9 

L  Z.o 

+  ,  etc. 

Ex.  1.  Find  the  12th  term  of  the  series  2,  6,  12,  20,  30,  etc. 
The  first  order  of  differences,       4,  6,  8,  10,  etc. 
"     second  order  of  differences,     2,  2,  2,      etc. 
"    third  order  of  differences,          0,  0. 
Here  D'=4,  D"=2,  and  D'"  =  0.     Also  a  =  2  and  n=  12. 
Hence  T13  =  2  +  11D'  +  55D"  =  2  +44  +  110  =  156,  Ans. 
Ex.  2.  Find  the  twentieth  term  of  the  series 

1,  3,  6,  10,  15,  21,  etc. 
Here  D'=2,  D/r  =  l,  a=l,  and  n  =  20. 
Therefore  T20=1-|-19D/+171D'/  =  1  +  38  +  171=210,  Ant. 
Ex.  3.  Find  the  thirteenth  term  of  the  series 

1,  5,  14,  30,  55,  91,  etc. 
Ex.  4.  Find  the  fifteenth  term  of  the  series 

1,  4,  9,  16,  25,  36,  etc. 
Ex.  5.  Find  the  twentieth  term  of  the  series 

1,  8,  27,  64,  125,  etc. 
Ex.  6.  Find  the  ftth  term  of  the  series  1,  3,  6,  10,  15,  21,  etc. 

rc 
Ans.  - 


262  ALGEBRA. 

Ex.  7.  Find  the  nth  term  of  the  series  1,  4,  10,  20,  35,  etc. 

Ans.  * 


Ex.  8.  Find  the  nth  term  of  the  series  1,  5,  15,  35,  70,  126,  etc, 

ttfoi+l)(n  +  2)(tt  +  3) 
24 

?/r 

371.  To  find  the  sum  of  n  terms  of  any  series: 
Let  us  assume  the  series 

0,  a,  a-f-&,  a+6-fc,  a-J-6+c+c?,  etc.       (1.) 
Subtracting  each  term  from  the  next  succeeding,  we  obtain 
the  first  order  of  differences, 

a,  bj  c,  d,  etc.  (2.) 

Now  it  is  clear  that  the  sum  of  n  terms  of  the  series  (2)  is 
equal  to  the  (n  +  l)th  term  of  series  (1);  and  the  nth  order  of 
differences  in  series  (2)  is  the  (n+l)th  order  in  series  (1).  If, 
then,  we  denote  the  sum  of  n  terms  of  series  (2)  by  S,  which  is 
the  same  as  the  (n-j-l)th  term  of  (1),  we  may  obtain  the  value 
of  S  from  the  formula  of  the  preceding  article  by  substituting 

0  for  a, 
tt  +  1  for  n, 
a  for  D', 
V  for  D",  etc. 
Hence 


etc. 


When  any  one  of  the  successive  orders  of  differences  be 
comes  zero,  this  formula  gives  the  exact  sum  of  the  terms. 
When  no  order  of  differences  becomes  zero,  the  formula  may 
still  give  approximate  results,  which  will,  in  general,  be  nearer 
the  truth  the  greater  the  number  of  terms  employed. 

EXAMPLES. 

1.  Find  the  sum  of  15  terms  of  the  series 

1,  3,  6,  10,  15,  21,  etc. 
Here  o=l,  D'  =  2,  D"  =  l,  D'"  =  0. 
Therefore     8=15-1-15.14+5.7.13=680,  Ana. 


SERIES.  263 


2.  Find  the  sum  of  20  terms  of  the  series 

1,  4,  10,  20,  35,  etc. 

3.  Find  the  sum  of  n  terms  of  the  series 

1,  2,  3,  4,  5,  6,  etc. 

A 
Ans. 

4.  Find  the  sum  of  n  terms  of  the  series 

I2,  22,  32,  42,  52,  etc. 


6 

5.  Find  the  sum  of  n  terms  of  the  series 
I3,  23,  33,  43,  53,  etc. 

( 


6.  Find  the  sum  of  n  terms  of  the  series 

1,  3,  6,  10,  15,  etc. 

Ans. 

7.  Find  the  sum  of  n  terms  of  the  series 

1.2,  2.3,  3.4,  4.5,  5.6,  etc. 

An8. 


8.  Find  the  sum  of  n  terms  of  the  series 
1,  4,  10,  20,  35,  etc. 


2.3.4 

Interpolation. 

372.  Interpolation  is  the  process  by  which,  when  we  have 
given  a  certain  number  of  terms  of  a  series,  we  compute  inter 
mediate  terms  which  conform  to  the  law  of  the  series. 

Interpolation  may,  in  most  cases,  be  effected  by  the  use  of 
the  formula  of  Art.  370.  If  in  this  formula  we  substitute  w  +  1 
for  n,  we  shall  have 

Tn+1=a+MD'+^D"+"(1\-2V+,  etc, 


which  expresses  the  value  of  that  term  of  the  series  which  has 
n  terms  before  it.     When  n  is  a  fraction  less  than  unity,  Trt+i 


264  ALGEBKA. 

stands  for  a  term  between  the  first  and  second  of  the  given 
terms.  When  n  is  greater  than  1  and  less  than  2,  the  inter 
mediate  term  will  lie  between  the  second  and  third  of  the  given 
terms,  and  so  on.  In  general,  the  preceding  formula  will  give 
the  value  of  such  intermediate  terms. 

EXAMPLES. 

1.  Given  the  cube  root  of  60  equal  to  3.914868, 

"  "  "       61       "        3.936497, 

"  "       62       "        3.957891, 

"  "  "       63       "        3.979057, 

"  "  "       64       "        4.000000, 

to  find  the  cube  root  of  60.25. 
Here  D'  =  +  .021629,  D"=-. 000235,  D//;=+. 000007,  etc 

a=3.914868,  and  w  =  .25. 
Substituting  the  value  of  n  in  the  formula,  we  have 

Tn+i^a+iD'-A^'+T^'''-,  etc. 
The  value  of  the  1st  term  is      +3.914868, 

"  "       2d         "  +  .005407, 

"  "       3d         «  +  .000022, 

"  "      4th       "  +  .000000. 

Hence  the  cube  root  of  60.25  is      3.920297. 

2.  Find  the  cube  root  of  60.5.  Ans.  3.925712. 

3.  Find  the  cube  root  of  60.75.  Ans.  3.931112. 

4.  Find  the  cube  root  of  60.6.  Ans.  3.927874. 

5.  Find  the  cube  root  of  60.33.  Ans.  3.922031. 

6.  Given  the  square  root  of  30  equal  to  5.477226, 

"  "  "          31  "       5.567764, 

«  "  "          32  "       5.656854, 

"  "  "          33  "       5.744563, 

«  "  "          34  "       5.830952, 

to  find  the  square  root  of  30.3  Ans.  5.504544, 

7.  Find  the  square  root  of  30.4.  Ans.  5.513619. 

8.  Find  the  square  root  of  30.5.  Ans.  5.522681. 

9.  Find  the  square  root  of  30.6.  Ans.  5.531727. 
10.  Find  the  square  root  of  30.8.  Ans.  5.549775. 


SERIES.  265 

Development  of  Algebraic  Expressions  into  Series. 

373.  An  irreducible  fraction  may  be  converted  into  an  in 
finite  series  by  dividing  the  numerator  by  the  denominator,  ac 
cording  to  the  usual  method  of  division. 

Ex.  1.  Expand  -  -  into  an  infinite  series. 
1  —  x 


x3+x*  +  1  etc. 

i-rflS 


X3 


Hence  -  --  =  1+  x-f  x2+#3+x4+x5-f  ,  etc.,  to  infinity. 
J.  —  x 

Suppose  x=^,  we  shall  then  have 


,  etc. 


Suppose  x—%,  we  shall  then  have 

,  etc. 


Ex.  2.  Convert  -  -  into  an  infinite  series. 
1-fo? 

Ans.  1—  x-\-x2—  x*-\-x*—  x54-,  etc. 
Suppose  03=4,  we  shall  then  have 

T|-|=i=l-i+i-H-rV-A+,  etc. 

Ex.  3.  Convert  -  into  an  infinite  series. 
a+x 

_.       XX2       X3      X4 

Ans.  1  ---  \-  —  —  -H  —  j—  ,  etc. 
a     a2     a3     a* 

Ex.  4.  Convert  -  into  an  infinite  series. 
a—x 

_,        X      X2      X3      X* 

Ans.  1+-  +—  +_4-_4-    etc. 
•-  a     a2     a3     «4 


266  ALGEBRA. 

1+x 

Ex.  5.  Convert  into  an  infinite  series. 

1— x 

Ex.  6.  Convert into  an  infinite  series. 

a — x 

,     2x     2x2  ,  2x3     2x*  , 
Ans.  1  +  — H — 2-+—+—+,  etc- 
a       a2       a3       a4 

Ex.  7.  Convert into  an  infinite  series. 

1— x+x2 

Ans.  1+x— x3— x4+x6+x7— ,  etc. 

1  — x 

Ex.  8.  Convert into  an  infinite  series. 

1— x+x2 

Ans.  1— x2— x3+x5+x6— x8— ,  etc. 

~1       i     /Y* 

Ex.  9.  Convert  = ^  into  an  infinite  series. 

1— x  — x2 


374.  An  algebraic  expression  which  is  not  a  perfect  square 
may  be  developed  into  an  infinite  series  by  extracting  its  square 
root  according  to  the  method  of  Art.  198. 

Ex.  1.  Develop  the  square  root  of  1  -fx  into  an  infinite  series. 
x    x2     x3     5x4 
2~8  +16-128+'  6tC- 


x 


x* 
4 

X2      X3   .    X4 


x3  x6 


5x4     x5       x6 


SERIES.  267 

Hence  the  square  root  of  1  -f  x  is  equal  to 


Suppose  a5=l,  we  shall  have 

V2  =  l+i—  J  +  Vo~  rfff  +  ,  etc. 
Ex.  2.  Develop  the  square  root  of  a2  +  x  into  an  infinite  series. 


Ex.  3.  Develop  the  square  root  of  a4  +x  into  an  infinite  series. 
Ex.  4.  Develop  the  square  root  of  a*—  x  into  an  infinite  series. 
Ex.5.  Develop  the  square  root  of  «2  +  x2  into  an  infinite 
series.  \ 

Method  of  Undetermined  Coefficients. 

375.  One  of  the  most  useful  methods  of  developing  algebraic 
expressions  into  series  is  the  method  of  undetermined  coefficients. 
It  consists  in  assuming  the  required  development  in  the  form 
of  a  series  with  unknown  coefficients,  and  afterward  finding  the 
value  of  these  coefficients.     This  method  is  founded  upon  the 
properties  of  identical  equations. 

376.  An  identical  equation  is  one  in  which  the  two  members 
are  identical,  or  may  be  reduced  to  identity  by  performing  the 
operations  indicated  in  them.     As 

ax  4-  b  =  ax  -f  5, 


a  —  x 

a          ax 

a—- =- . 

1+x     l+x 

377.  It  follows  from  the  definition  that  an  identical  equation 
is  satisfied  by  each  and  every  value  which  may  be  assigned  to  a  let 
ter  which  it  contains,  provided  that  value  is  the  same  in  both 
members  of  the  equation. 

Every  identical  equation  containing  but  one  unknown  quan 
tity  can  be  reduced  to  the  form  of 

3-K  etc. 


268  ALGEBRA. 

378.  If  an  equation  of  the  form 

A  +  Bx  +  Cx2  +  ,  etc.  = 
must  be  satisfied  for  each  and  every  value  given  to  x,  then  the  co 
efficients  of  the  like  powers  of  x  in  the  two  members  are  equal  each 
to  each. 

For,  since  this  equation  must  be  satisfied  for  every  value  of 
ic,  it  must  be  satisfied  when  x  =  0.  But  upon  this  supposition 
all  the  terms  vanish  except  two,  and  we  have 

A  =  A'. 
Suppressing  these  two  equal  terms,  we  have 

Bx  +  Cx2  +  ,  etc.=B/x  +  C'x2  +  J  etc. 
Dividing  each  term  by  x,  we  obtain 

B  +  Cx  +  ,  etc.  =  B/  +  C'x  +  ,  etc. 

Since  this  equation  must  be  satisfied  for  every  value  of  x,  it 
must  be  satisfied  when  x=0.  But  upon  this  supposition 

B  =  B'. 

In  the  same  manner  we  can  prove  that 

0=0', 
D=D',  etc. 

379.  Whenever  we  have  an  equation  of  the  form 

M  +  Nx  +  Px2  +  Qx3  +  ,  etc.  =  0, 

which  is  true  for  every  value  of  x,  all  the  coefficients  ofx  are  equal 
to  zero. 

For,  if  we  transpose  all  the  terms  of  the  equation  in  the  last 
article  to  the  left-hand  member  we  shall  have 


But  it  has  been  shown  that  A  =  A',  B  =  B',  etc.-  whence 
A—  A'  =  0,  B—  B'  =  0,  etc.  If  we  substitute  M  for  A  —  A',  and 
N  for  B  —  B',  etc.,  the  equation  will  be 

M  +  Nx  +  Px24-Qx3  +  ,  etc.  =  0. 
whence  M-0  N  =  0  P  =  0  etc. 


- 

Ex.  1.  Expand  the  fraction  ^-^  —  into  an  infinite  series. 

1  —  3x 

It  is  plain  that  this  development  is  possible,  for  we  may  divide 
the  numerator  by  the  denominator,  as  explained  in  Art  373. 
Let  us,  then,  assume  the  identical  equation 


SERIES.  269 


etc., 


where  the  coefficients  A,  B,  C,  D  are  supposed  to  be  independ 
ent  of  x,  but  dependent  on  the  known  terms  of  the  fraction. 

In  order  to  obtain  the  values  of  these  coefficients,  let  us  clear 
this  equation  of  fractions,  and  we  shall  have 


(E-3D)x4+,  etc. 

Now,  since  this  is  supposed  to  be  an  identical  equation,  the 
coefficients  of  the  like  powers  of  x  in  the  two  members  are 
equal  each  to  each. 
Therefore  A  =  l. 

B-3A  =  2,  whence  B=5; 
C-3B  =  0,         "       0  =  15; 
D-3C  =  0,         "      D=45; 
E-3D  =  0,         "      E  =  135,  etc. 

Substituting  these  values  of  the  coefficients  in  the  assumed 
series,  we  obtain 


where  the  coefficient  of  each  term  after  the  second  is  three 
times  the  coefficient  of  the  preceding  term. 

380.  The  method  thus  exemplified  is  expressed  in  the  fol 
lowing 

RULE. 

Assume  the  proposed  expression  equal  to  a  series  of  the  form 
A  +  Bx+C^2  +  ,  etc.  ;  clear  the  equation  of  fractions,  or  raise  it  to 
its  proper  power,  and  place  the  coefficients  of  the  like  powers  of  x  in 
the  two  members  equal  each  to  each.  Then  find  from  these  equa 
tions  the  values  of  A,  B,  C,  etc.,  and  substitute  these  values  in  the, 
assumed  development. 

Ex.  2.  Expand  the  fraction  -  into  an  infinite  series. 

1  —  2X+X2 

Assume  -  —  —  -  -  =  A  +  Bx+Cx2  +  Dcc3+Ex4+,  etc. 

JL  •  —  ^u^L*  ~Y~  OC 

Clearing  of  fractions,  we  have 


270  ALGEBRA. 

,  etc. 


Therefore  we  must  have 
A«lf 
B-2A  =  0,  whence  B  =  2 


:,  z=-=5,  etc. 

Therefore     _  =l  +  2as+3s8  +  4a?  +  5a4  +  ,  etc. 


Ex.  3.  Expand  the  fraction       +  X-  into  an  infinite  series, 

l  —  x— x2 

where  the  coefficient  of  each  term  is  equal  to  the  sum  of  the 
coefficients  of  the  two  preceding  terms. 

Ex.  4.  Expand  - —  ~X     2  into  an  infinite  series. 

J.  ~~~"  £OC  -~~  Ot£ 

What  is  the  law  of  the  coefficients  in  this  series? 

Ex.  5.  Expand  Vl  —  x  into  an  infinite  series. 

x    x2     x3     5cc4 
'2~8"~l6~~128~~256 


x    x2     x3     5cc4      7x5 
Ans.  1-  -        -~       etc. 


1  —  x 

Ex.  6.  Expand  •=  -  5  into  an  infinite  series. 
1xx2 


Ans.  1  —  2x+x2+x3  —  2x4+x5-j-o;6—  ,  etc. 
Ex.  7.  Expand  Va2—  x2  into  an  infinite  series. 

381.  Proper  Form  of  the  assumed  Series.  —  In  applying  the 
method  of  undetermined  coefficients  to  develop  algebraic  ex 
pressions  into  series,  we  should  determine  what  power  of  the 
variable  will  be  contained  in  the  first  term  of  the  development, 
and  assume  a  corresponding  series  of  terms.  Generally  the 
first  term  of  the  development  is  constant,  or  contains  x0'  but 
the  first  term  of  the  series  may  contain  x  with  any  exponent 
either  positive  or  negative.  If  the  assumed  development  com 
mences  with  a  power  of  x  lower  than  is  necessary,  no  error  will 


SEKIES.  271 

result,  for  the  coefficients  of  the  redundant  terms  will  reduce 
to  zero.  But  if  the  assumed  development  commences  with  a 
power  of  x  higher  than  it  should,  the  fact  will  be  indicated  by 
an  absurdity  in  one  of  the  resulting  equations. 

The  form  of  the  series  which  should  be  adopted  in  each  case 
may  be  determined  by  putting  x=0,  and  observing  the  nature 
of  the  result.  If  in  this  case  the  proposed  expression  becomes 
equal  to  a  finite  quantity,  the  first  term  of  the  series  will  not 
contain  x.  If  the  expression  reduces  to  zero,  the  first  term  will 

A 

contain  x  ;  and  if  the  expression  reduces  to  the  form  —  ,  then 

the  first  term  of  the  development  must  contain  x  with  a  nega 
tive  exponent. 

Let  it  be  required  to  develop        —  5  into  a  series. 
Assume       —  -  -=A+~Bx+ 

OX  —  X 

Clearing  of  fractions,  we  have 


etc., 

whence,  according  to  Art.  378,  we  obtain  1  =  0,  which  is  ab 
surd,  and  shows  that  the  assumed  form  is  not  applicable  in  the 
present  case. 

Let  us,  however,  assume 

etc. 


Clearing  of  fractions,  we  have 

l  =  3A+(3B-A)x+(3C-B)x2+(3D-C)#3+,  etc. 
Therefore  3A  =  1,  whence  A  =  £; 


Substituting  these  values,  we  find 

/•yi  — 1          /y&  /-v»  /y»2 

I i       * '      \          '        \          rt-f  /> 

o^     ^2 ~""~Q~"t~'q"i~97r~'  ftT"1"'  eic* 

OX  —  X  O  t/        ^i  I        OX 


272  ALGEBRA. 

To  Resolve,  a  Fraction  into  Simpler  Fractions. 

382.  When  the  denominator  of  a  fraction  can  be  resolved 
into  factors,  the  principles  now  developed  enable  us  to  resolve 
the  fraction  itself  into  two  or  more  simpler  fractions,  having  these 
factors  for  denominators.  In  such  a  case,  the  given  fraction  is 
the  sum  of  the  partial  fractions. 

Ex.  1.  Eesolve  the  fraction  -= — into  partial  fractions. 

x2— 5x  +  6 

We  perceive  that  x2-5x+6  =  (x-2)(cc-3). 

.                            5x-12         A         B 
Assume  -= — = — —^  = -+• 


x2—5x+6    x—2    x—  3' 
in  which  the  values  of  A  and  B  are  to  be  determined. 
Clearing  of  fractions,  we  have 


By  the  principle  of  Art.  378,  A  +  B=5, 
and  3A  +  2B  =  12. 

From  which  we  obtain  A  =2  and  B  =  3. 

Substituting  in  the  assumed  equation,  we  have 

5a?-12          2          3 
#2_5x+6~x—  2     a?—  3' 

5x-i-  1 

Ex.  2.  Eesolve  —  —  T  into  partial  fractions. 
x2—  1 

2 
An*. 


07+1       X— 
£)£,  _  ^Q 

Ex.  3.  Eesolve  —  —  ^  -  ^  ^nto  Partia^  fractions. 

3          2 
Ans.  --  -H 


. 

x  —  o     x  —  o 

3x2—  1 
Ex.  4.  Eesolve  —  ^  --  into  partial  fractions. 

00   —  —  OC 

Ans.  —  —  r-H 


—  —  -  , 

cc-f-1     x—  1     x 

K    ~      ,  2x2—  6x-f  6  ......       . 

Ex.  5.  Eesolve  7  -  —7  —  —  ^  into  partial  fractions. 

(x-l)(x-'2)(x-3) 

1  2  3 

Ans.  - 


-2     x-3° 


SERIES. 


278 


into  partial  fractions. 


Ex.  6.  Resolve 


Ex.  7.  Eesolve 


Reversion  of  Series. 

383.  ^7ie  reversion  of  a  series  is  the  finding  the  value  of  the 
unknown  quantity  contained  in  an  infinite  series  by  means  of 
another  series  involving  the  powers  of  some  other  quantity. 

This  may  be  accomplished  by  the  method  of  undetermined 
coefficients  in  a  mode  similar  to  that  employed  in  Art.  379. 

Ex.  1.  Given  the  series  y—x+x2-\-x*+,  etc.,  to  find  the  value 
of  x  in  terms  of  y. 

Assume          x=Ay+Vy*  +  Cy3  +  Dy*  +  .  etc. 

Find,  by  involution,  the  values  of  x2,  x3,  x4,  and  x5,  carrying 
each  result  only  to  the  term  containing  y5.  Then,  substituting 
these  values  for  x,  cc2,  x3,  etc.,  in  the  given  equation,  we  shall  have 


+2AB 
+  A3 


__J-2AC 

HB 

+  3^ 

+  A4 


+  2AD 

+  2BC 


+  3AB2 
+4A3B 


Since  this  is  an  identical  equation,  we  place  the  coefficients 
of  the  like  powers  of  y  in  the  two  members  equal  to  each  other, 
and  we  obtain 

A=+l,  B=-l,  C=+l,  D=-l,  E=+l,  etc. 

Hence  we  have  x=y—y2  +  y*  —  if-}-y5  —  )  etc.,  Ans. 


Ex.  2.  Given  the  series  y=x—  —  -f- — -5-  +  ,  etc.,  to  find  the 

A  <±  O 

value  of  x  in  terms  of  y.  2     ?,s     ?  4 

^4.7Z5.  x=7/+^-+4-+^4-,  etc. 

M  ? 


274  ALGEBRA. 

/y>2  ,y>3  ^4  /Y" 

Ex.  3.  Given  the  series  y=oc—  —  -{-—  ——-}-—  —  ^  etc.,  to  find 

,4       o      4      o 

the  value  of  x  in  terms  of  y. 

Ans.  x=y  +  1l  +  j^  +  ^+^f^>  +  ,  etc. 

Ex.  4.  Given  the  series  i/—x-\-x3-\-x5-\-x1  -\-x*-\-,  etc.,  to  find 
the  value  of  x  in  terms  of  y. 

Ans.  x  —  y  —  ?/3  +  2?/5  —  5?/7-fl4?/9  —  ,  etc. 

Ex.  5.  Given  the  series  2/=^  +  3x2+5o;3  +  7a?4H-9x5  +  ,  etc..  to 
find  the  value  of  x  in  terms  of  y. 

Ans.  x=y—  3/+13?/3—  67/4-381?/5—  ,  etc. 

384.  When  the  sum  of  a  series  is  known,  we  may  sometimes 
obtain  the  approximate  value  of  the  unknown  quantity  by  re 
verting  the  series. 

Ex.1.  Given  ^+ixH^3  +  ^ir^4  +  ;r«V<j^5  +  ,  etc.=l,  to 
find  the  value  of  x. 

If  we  call  s  the  sum  of  the  series,  and  proceed  as  in  the  last 
article,  we  shall  have 

X=2s-s2+fs3--i-s44-fs5-,  etc. 

Substituting  the  value  of  s,  we  find 

03  =  4—  -rV+  y-V—  imr+Wg-o  —  ,  etc.,  or  x  =  0.446354  nearly. 

Ex.  2.  Given  2x  +  3x3  +  4x5  +  5x7  +  ,  etc.=  J,  to  find  the  value 

ofx-  5     353     1955     152s7 


or        a=J-Tf*  +  T^-TBy»rT+,  etc.=:0.2300  nearly. 

^3        O"          o^ 

Ex.3.  Given  .T—  —  +———  +  ,  etc.  =^,  to  find  the  value  of  x. 
o      o       < 


)., 
or       a?=t+A-+Tnrer+  asUo6  +  ,  etc.  =  .34625  nearly. 


/yi2  /y>3  O^  'T1 

Ex.  4.  Given  x-\-  —  -\-  —  -}-  —  +  —  -{-^  etc.=-J-,  to  find  the  value 

Jj  O  4:          O 


ofa;- 


, 
=s__  +  ___  +  ___  +  i  etc., 


Or  33  =  "5  -  15  7T  ~T~  7  5  0         '1  A  0001~375000  -  1125000o~i~7 

or         a  =  0.1812692  nearly. 


SERIES.  275 

Binomial  Theorem. 

385.  In  Art.  353  the  binomial  theorem  was  demonstrated 
for  the  case  in  which  m  is  a  positive  whole  number.    By  means 
of  the  method  of  undetermined  coefficients  we  can  prove  that 
this  formula  is  true,  whether  m  is  positive  or  negative,  entire  or 
fractional.     The  demonstration  of  this  theorem  depends  upon 
the  following  proposition : 

(yn Jn 

386.  The  value  of  -    — — ,  when  a— &,  is  in  all  cases  nan~l, 

whether  n  is  positive  or  negative,  integral  or  fractional. 

first.  It  was  shown  in  Art.  83  that  when  n  is  a  positive 
whole  number,  an—bn  is  exactly  divisible  by  a  —  b,  and  the 

quotient  is  an-l  +  an-*b  +  an-*bz+ +6n~1.    The  number  of 

terms  in  this  quotient  is  equal  to  n;  for  b  is  contained  in  all  the 
terms  except  the  first,  and  the  exponents  of  b  are  1,  2,  3,  etc., 
to  n— 1,  so  that  the  number  of  terms  containing  b  is  n— 1,  and 
the  whole  number  of  terms  is  equal  to  n.  Now,  when  a—b, 
each  term  of  the  above  quotient  becomes  a71"1,  and,  since  there 
are  n  terms  in  the  quotient,  this  quotient  reduces  to  nan~l. 

Second.  Suppose  n  to  be  a  positive  fraction,  or  n=—,  where 

p  and  q  are  positive  whole  numbers. 

1  P 

Let  ai=x,  whence  aq=xp,  and  a=xq. 

i  p 

Also,  let  &=y,  whence  b^—yP,  and  b  =  yQ. 

Then,  substituting,  we  have 

P      P  xP-yP 

an—bn     aq  —  bi     xp—yp        x—y 
a_&  ~~  a— b   ~  x<i—yi~  afl—y* 

x-y 

But  p  and  q  are  positive  integers;  therefore,  when  a =b,  and, 
consequently,  x=y,  according  to  case  first,  the  numerator  of 
the  last  fraction  becomes  pxp~l,  and  the  denominator  becomes 
qx*-1 ;  that  is,  the  fraction  reduces  to 

pxP~l        p 

r,  or  —xP-<i. 

qx<i-1          q 


276  ALGEBRA. 

1 

Substituting  for  x  its  value  afl,  the  fraction  reduces  to 

p    £Z«  p    *_.! 

—  a  v  .or  -a«     .or  «a*~*. 
2  2 

Third.  Suppose  n  to  be  negative,  and  either  integral  or  frac« 
tional  ;  or  let  n——m.     Then  we  shall  have 

I  __  l_ 

an—bn_a-m—b-m_a™~'b™__    1      bm—am  I      am—bm 

a—b  ~       a—b        '    a—b   ~  ambm'    a—-b   ~    ~a"'bm'    a  —  b  ' 
Now,  when  a—b^  the  first  factor  of  the  last  expression  re 

duces  to  --  ^,  or  —  a~2m,  and  the  second  factor  (by  one  of  the 
a 

preceding  cases)  reduces  to  mam~l.     Hence  the  expression  be 
comes  —ar*mXmaP-'\  or  —  ma-™-1,  QTnan~l. 

387.  It  is  required  to  obtain  a  general  formula  expressing  the 
value  of(x  +  a)m,  whether  m  be  positive  or  negative,  integral  or 

fractional. 

(a\  /       a\m 

1  +  -);  therefore  (x+a)m=xm(l-\--)  . 

(a\m 
1+-)  ,  we  have 

only  to  multiply  it  by  xm  to  obtain  that  of  (x+a)m. 


Let  -=z;  then,  to  develop  (l  +  z)"1,  assume 

(l  +  2)™^A  +  Bs+O;2+TV+,  etc.,  (1.) 

in  which  A,  B,  C,  D,  etc.,  are  coefficients  independent  of  z,  and 
we  are  to  determine  their  values, 

Now  this  equation  must  be  true  for  any  value  of  z;  it  must 
therefore  be  true  when  2  =  0,  in  which  case  A  =  l. 
Substituting  this  value  of  A  in  Eq.  (1),  it  becomes 

(1  +  2)*»  =  1  +  Bz  +  Cz2  +  Dz2  +  ,  etc.  (2.) 

Since  Eq.  (2)  is  to  be  true  for  all  values  of  z,  let  z=n  •  then 
(2)  becomes 

(l  +  ?2)m=1  +  B7^^-(>^2  +  D?^3  +  ,  etc.  (3.) 

Subtracting  (3)  from  (2),  member  from  member,  we  have 


SERIES.  277 

Dividing  the  first  member  of  (4)  by  (1  +  z)  —  (1  +  n)7  and  the 
second  by  its  equal  z—  n,  we  have 


z—n          z  —  n 

But  when  z=n,  or  1  +  2  =  1  +n,  the  first  member  of  equation 
(5)  becomes  m(l  +  z)711-1. 

g2  _  72,2 

Also,  —  =«  +  n,  when  s=w,  becomes  2z. 

2  —  ?1 
Z3  —  ft3 

—  —  z2+zn-\-n2,  when  3=n,  becomes  3z2,  etc, 

£  —"•  Tc- 

These  values  substituted  in  (5)  give 

ra(l-|-s)m-1=:B+2Cs+3Ds2  +  4Es3  +  ,  etc<  (6>j 

Multiplying  both  members  of  Eq.  (6)  by  1  +  z,  we  have 
m(l  +  2)-=B  +  (2C  +  B)s  +  (3D  +  2C>2+(4:E  +  3D>3  +  3etc.  (7.) 
If  we  multiply  Eq.  (2)  by  m,  we  have 

m(l  +  s)m  =  m  +  ?7^Bz  +  mCs2  +  mD23^-,  etc.  (8.) 

The  first  members  of  Eq.  (7)  and  (8)  are  equal;  hence  their 
second  members  are  also  equal,  and  we  have 

-,  etc.— 

etc.        (9.) 

This  equation  is  an  identical  equation  ;  that  is,  it  is  true  for 
all  values  of  z.  Therefore  the  coefficients  of  the  like  powers 
of  z  in  the  two  members  are  equal  each  to  each,  and  we  have 

B=fli. 

2C+B=wB,  whence  C  =  m^"^; 

2t 

^-l)(!!!=2)t  etc. 

^.O 

Substituting  these  values  in  (2),  we  have 
(l+^  =  l  +  m.+^-^^  +  ro(i"-y'-2^  +  ,etC.  (10.) 

If  in  this  equation  we  restore  the  value  of  z,  which  is  ^,  we 

have 

a     m(m  —  l)  a2     m(m  -!)(???,  -2)   a3 

a; 


-  .  _.         _  -         +  etc.; 

xj  x  2          ,x2  2.3  3 


278  ALGEBRA. 

and  multiplying  both  members  by  xm,  we  obtain 

o-io  I  <rv)    _      ~I  1 

(x  +  a)m  =  x 


-'a'  +  ,etc,  (11.) 

which  is  the  general  formula  for  the  development  of  any  bino 
mial  (x  +  a)m,  whatever  be  the  values  of  x  and  a,  and  whether 
m  be  positive  or  negative,  integral  or  fractional  ;  and  this  for 
mula  is  known  as  the  Binomial  Theorem  of  Sir  Isaac  Newton. 

388.  When  the  Series  is  Finite.  —  The  preceding  development 
is  a  series  of  an  infinite  number  of  terms  ;  but  when  m  is  a  pos 
itive  integer,  the  series  will  terminate  at  the  (??z  +  l)th  term, 
and  all  the  succeeding  terms  will  become  zero.     For  the  second 
term  of  Eq.  (11)  contains  the  factor  m,  the  third  term  the  factor 
m—  1,  the  fourth  term  the  factor  m  —  2,  and  the  (m  +  2)d  term 
contains  the  factor  m  —  m,  or  0,  which  reduces  that  term  to  0; 
and  since  all  the  succeeding  terms  also  contain  the  same  factor, 
they  also  become  0.     There  will  therefore  remain  only  m-f-1 
terms. 

When  m  is  not  a  positive  integer,  it  is  evident  that  no  one 
of  the  factors  m,  m—  1,  m  —  2,  m—  3,  etc.,  can  be  equal  to  0, 
so  that  in  that  case  the  development  will  be  an  infinite  series. 

389.  Expansion  of  Binomials  with  negative  integral  Exponents 
This  is  effected  by  substitution  in  formula  (11). 

Ex.  1.  Expand  -  -  or  (a  +  b}~1  into  an  infinite  series. 
a  +  b 

In  (11)  let  ra=  —  1,  and  we  find 
the  coefficient  of  the  second  term  is  —  1, 

"  "       third        "     is  •~1*~2=:+1, 

A 

"  "      fourth      "    is  ~ 


, 

o 

"       fifth          "     is  ~1*~"4=+1,  etc. 


SERIES.  279 

Hence  we  have 


,  etc., 
1     _1     IV*     &_3 

M^~«~«2     a3~«4 

which  is  an  infinite  series,  and  the  law  of  the  series  is  obvious. 
We  might  have  obtained  the  same  result  by  the  ordinary  meth 
od  of  division. 

Ex.  2.  Expand  -f  -  =—  or  (a  +  6)~2  into  an  infinite  series. 
(a+Vf 

Ans.  a-2-2a-364-3a-462-4a-5£34-5a-664-,  etc., 


or  -5  --  H  —  r  --  r  +  ~6~~;  e^c-> 

a2     a3      a4      a5      a6 

where  the  law  of  the  series  is  obvious. 

Ex.  3.  Expand  --  j  or  (a—b)-1  into  an  infinite  series. 

Ans.  a-1  +  a-*b  +  a-*bz     a-4b3  +    etc. 


Ex.  4.  Expand  -, r—  or  (a—b)~2  into  an  infinite  series. 

(a  —  b)2 

Ex.  5.  Expand  (a-f&)~3  into  an  infinite  series. 

Ans.  a~3  —  Sa~~4b  +  6a~5b2 — Wa~Gb3  +  15a~7^4 — ,  etc. 
Ex.  6.  Expand  (a  —  b)-4  into  an  infinite  series. 

Ex.  7.  Expand  (i  +  2x)~5  into  an  infinite  series. 

Ans.  1  — 10x-}-60x2-280x3+,  etc. 

390.  Expansion  of  Binomials  with  positive  Fractional  Exponents. 

Ex.  1.  Expand  Va-\-b  or  (a+6)2  into  an  infinite  series. 
Represent  the  coefficients  of  the  different  terms  by  A,  B,  Q 
D,  etc. ;  then 

A=  +1, 

B=         n       =+*. 


2.4' 


3  2.4.6' 

?7-3  1.3.5 


280  ALGEBRA. 

Hence  we  have 

JLJL1JL  1  3  10  5  1     Q    P;  7 

(7  \  2  2  -1-       ~~  2  7  -*"  *?70  -L  .  O  ~1F7  o  i»O»U  --  5~74 

a+^a'+-  2*-«      2 


+  ,  etc. 

The  factors  which  form  the  coefficients  are  kept  distinct,  in 
order  to  show  more  clearly  the  law  of  the  series.  The  numer- 
ators  of  the  coefficients  contain  the  series  of  odd  numbers,  1,  3, 
5,  7,  etc.,  while  the  denominators  contain  the  even  numbers, 
2,  4,  6,  8,  etc. 

Ex.  2.  Expand  (x—  a)2  into  an  infinite  series. 

Ex.  3.  Expand  (a2  +  x)3  into  an  infinite  series. 

,  a-^x     a-V     3«-5x3     3.5a~V 
An*.  a+—  __-4.____2T_nr_  +  ,  etc. 

x         x2  Sx3  3.5x4 

a  +      "  +  -ete- 


Ex.  4.  Expand  (a-f-5)*  into  an  infinite  series. 

b        W         2.563          2.5.8i4 


Ex.  5.  Expand  (a3—  63)8  into  an  infinite  series. 

2.5/>9 


, 

AnS.  a  1  1  —  —  -— 


fl    6o    g    n    Q> 

3a3     3.6ci6     3.6.9a9 

_2_ 

Ex.  6.  Expand  (a+^c)8  into  an  infinite  series. 
Ex.  7.  Expand  (a—b)^  into  an  infinite  series. 

ij-i     A       3^2          3.7£3  3.7.1164 

~ 


2     4.8.12a3~4.8.12.16a4     ' 
Ex.  8.  Expand  (1—  x}5  into  an  infinite  series. 


x      4x2        4.9x3         4.. 
~5~630~5.10.15~5.10.15.20~'  C 

J|L 

>F 
391.  Expansion  of  Binomials  ivith  negative  Fractional  Expo 

nents. 

Ex.  1.  Expand  -      —^  or  (a  +  b)   a  into  an  infinite  series. 


SERIES.  281 

The  terms  without  the  coefficients  are 

a~*  a~h,  a~h*,  a~~h3,  a~V,  etc. 
Kepresent  the  coefficients  by  A,  B,  C,  D,  etc. ;  then 
A-  +1, 

B-        n       =  - 


n-cl_     1.3.5 
~1T~~274~6' 


Hence  we  obtain 

-4       -i    1  -|7  ,  1.3  -|79     1.3.5  -$73  ,  1.3.5.7  -S 
=a     -ga     b  +  -a  ^-jf^  ^+0^   ~ 

—  ,  etc. 

ft        8&a         3.5Z>3 
+ 


2.4.6a32.4.6.8a4     '* 

Ex.  2.  Expand  (a2—  #)~2  into  an  infinite  series. 

1  ,    a;    ,  1.3a;2  ,  1.3.  ox3  ,  1.3.6.7a^ 

S-  a+2a3  +  2.4a5+2.4.6a7+^4r6.8a9  +  '  6 

97? 

Ex.  3.  Expand     .  into  an  infinite  series. 

Vx2+a4 

A       m  j-t      a4   .  l-3«8     1.3.5ft12     1.3.5.7a16 
?'         1          +  + 


Ex.4.  Expand  (a-fx)   *  into  an  infinite  series. 

_i     1  _|       1.4  -f  2     1  .  4  .  7  -L°  3     1.4.7.10  -J.J3   4 
~3a  3.6a         ~3 . 6 .  9a       X      3. 679712°       'C    •' 

Ex.  5.  Expand  (a2— x2)  *  into  an  infinite  series. 


1    J-,  ,   x2   ,  1.5x4      1.5.9x6  ) 

AnS'  V*  \  1+4^+4^+OTl2^  +  '  6ta  f 


Ex.  6.  Expand  (1+x)   5  into  an  infinite  series. 
x      6x2        6.11x3        G.ll.i 


282  ALGEBKA. 

392.  Extraction  of  any  Root  of  a  Surd  Number. — The  approx 
imate  value  of  a  surd  root  may  be  found  by  the  binomial 
theorem  by  dividing  the  number  into  two  parts,  and  consider 
ing  it  as  a  binomial. 

Ex.  1.  Find  the  square  root  of  10. 


second     " 

+   .1666667 

third 

-   .0046296 

fourth      " 

4-  .0002572 

fifth 

-  .0000179 

sixth        " 

+  .0000014 

seventh  " 

-  .0000001 

If,  in  Ex.  3,  Art.  390,  we  make  a2  =  9  and  as=l,  we  shall  have 

1  -I  o  QfC  Q      £i     7 

27~3~2.4.33     2.4.6.3b~2.  4  .  6  .  8  .37+2.4.6.8.10.  39~'  6tC* 

The  value  of  the  first  term  is     3.0000000 

u 
u 

u 
tt 
u 

Their  sum  is  3.1622777, 

which  is  the  square  root  of  10  correct  to  seven  decimal  places. 
Ex.  2.  Find  the  square  root  of  99. 

-/99=  VlOO— 1  =  (100— 1)  . 
Substituting  in  Ex.  3,  Art.  390,  we  have 

V99  =  10 — • — — t  5— ,  etc. 

The  value  of  the  first  term  is     10.0000000 

"  "      second     "    -     .0500000 

"      third        "    -     .0001250 

"  "      fourth      "    -     .0000006 


Their  sum  is  9.9498744, 

which  is  the  square  root  of  99  correct  to  seven  decimal  places. 

393.  The  method  here  exemplified  for  finding  the  nth  root 
of  any  number  is  expressed  in  the  following 

RULE. 

Find,  by  trial,  the  nearest  integral  root  (a),  and  divide  the  given 
number  into  two  parts,  one  of  which  is  the  nth  power  of  (a).     Con- 


SERIES.  283 

sider  these  two  parts  as  the  terms  of  a  binomial,  and  develop  it  into 
a  series  by  the  binomial  theorem. 

Ex.  3.  Find  the  cube  root  of  9  to  seven  decimal  places. 

1  2  2.5  2.5.8 

+       ^~+"«~ll"h'  e 


=  2.0800838. 

Ex.  4.  Find  the  cube  root  of  31  to  seven  decimal  places. 
A      qli        4          2.42          2.5.43          2.5.S.44  1 

r  +  3.27~3.6.272  +  3.6.9.273~3.6.9.12.274+'       ') 

=  3.1413806. 
Ex.  5.  Find  the  fifth  root  of  30  to  seven  decimal  places. 

2  2.4  2.4.9 

5.16     5.10.162    5.10.15.163~'e: 
=  1.9743506. 

^ 


234  ALGEBKA. 


CHAPTEE  XX. 

LOGARITHMS. 

394.  The  logarithm  of  a  number  is  the  exponent  of  the  power 
to  which  a  constant  number  must  be  raised  in  order  to  be  equal 
to  the  proposed  number.     The  constant  number  is  called  the 
base  of  the  system. 

Thus,  if  a  denote  any  positive  number  except  unity,  and 
a2=m,  then  2  is  the  exponent  of  the  power  to  which  a  must 
be  raised  to  equal  m;  that  is,  2  is  the  logarithm  of  m  in  the 
system  whose  base  is  a.  If  a*=ra,  then  x  is  the  logarithm  of 
m  in  the  system  whose  base  is  a. 

395.  If  we  suppose  a  to  remain  constant  while  m  assumes  in 
succession  every  value  from  zero  to  infinity,  the  corresponding 
values  of  x  will  constitute  a  system  of  logarithms. 

Since  an  indefinite  number  of  different  values  may  be  attrib 
uted  to  a,  it  follows  that  there  may  be  an  indefinite  number  of  sys 
tems  of  logarithms.  Only  two  systems,  however,  have  come  into 
general  use,  viz.,  that  system  whose  base  is  10,  called  Briggs's 
system,  or  the  common  system  of  logarithms ;  and  that  system 
whose  base  is  2.718 +  ,  called  the  Naperian  system,  or  hyperbolic 
system  of  logarithms. 

Properties  of  Logarithms  in  general. 

396.  TJie  logarithm  of  the  product  of  two  or  more  numbers  is 
equal  to  the  sum  of  the  logarithms  of  those  numbers. 

Let  a  denote  the  base  of  the  system ;  also,  let  m  and  n  be 
any  two  numbers,  and  x  and  y  their  logarithms.  Then,  by  the 
definition  of  logarithms,  we  have 

a*=m,  (1.) 

a.v=n.  (2.) 

Multiplying  together  equations  (1)  and  (2)  member  by  mem 
ber,  we  have  a*+y=mn. 


LOGARITHMS.  285 

Therefore,  according  to  the  definition  of  logarithms,  x+y  is 
the  logarithm  of  mn,  since  it  is  the  exponent  of  that  power  of 
the  base  which  is  equal  to  mn. 

For  convenience,  we  will  use  log.  to  denote  logarithm,  and 
we  have 

x  +  y  =  log.  mn  =  log.  m  +  log.  n. 

Hence  we  see  that  if  it  is  required  to  multiply  two  or  more 
numbers  together,  we  have  only  to  take  their  logarithms  from 
a  table  and  add  them  together;  then  find  the  number  corre 
sponding  to  the  resulting  logarithm,  and  it  will  be  the  product 
required. 

397.  The  logarithm  of  the  quotient  of  two  numbers  is  equal  to 
the  logarithm  of  the  dividend  diminished  by  that  of  the  divisor. 
If  we  divide  Eq.  (1)  by  Eq.  (2),  member  by  member,  we  shall 

m 

have  ax~v=—. 

n 

Therefore,  according  to  the  definition,  x — y  is  the  logarithm 

777 

of  — ,  since  it  is  the  exponent  of  that  power  of  the  base  a  which 
is  equal  to  — .     That  is, 


n 
x 


—  y  =  log.  f~-l=~]pg.  m— log.  n. 


Hence  we  see  that  if  we  wish  to  divide  one  number  by  an 
other,  -we  have  only  to  take  their  logarithms  from  the  table  and 
subtract  the  logarithm  of  the  divisor  from  that  of  the  dividend ; 
then  find  the  number  corresponding  to  the  resulting  logarithm, 
and  it  will  be  the  quotient  required. 

398.  The  logarithm  of  any  power  of  a  number  is  equal  to  the 
logarithm  of  that  number  multiplied  by  the  exponent  of  the  power. 

If  we  raise  both  members  of  Eq.  (1)  to  any  power  denoted 
by  Pi  we  have  apx=mp. 

Therefore,  according  to  the  definition,  px  is  the  logarithm  of 
mp,  since  it  is  the  exponent  of  that  power  of  the  base  which  is 
equal  to  mP.  That  is, 

—p  log.  m. 


286  ALGEBRA. 

Therefore,  to  involve  a  given  number  to  any  power,  we 
multiply  the  logarithm  of  the  number  by  the  exponent  of  the 
power;  the  product  is  the  logarithm  of  the  required  power. 

399.  The  logarithm  of  any  root  of  a  number  is  equal  to  the  log 
arithm  of  that  number  divided  by  the  index  of  the  root 

If  we  extract  the  rth  root  of  both  members  of  Eq.  (1),  we 

X 

shall  have  ar=  y/ra. 

Therefore,  according  to  the  definition,  -  is  the  logarithm  of 

l/ra.     That  is 

x     ,       r  i —     loo;,  m 
-—log.  \/m=— — . 


r  r 


Therefore,  to  extract  any  root  of  a  number,  we  divide  the 
logarithm  of  the  number  by  the  index  of  the  root  ;  the  quotient 
is  the  logarithm  of  the  required  root. 

400.  The  following  examples  will  show  the  application  of 
the  preceding  principles  : 

Ex.  1.  log.  (abcd)=log.  a  +  log.  &-|-log.  c-f-log.  d. 

Ex.  2.  log.  f^-J  =log.  a  +  log.  6  +  log.  c—  log.  d—  log.  e. 

Ex.  3.  log.  (ambncp)  —  m  log.  a  +  n  log.  b+p  log.  c. 

(a?n£n\ 
—  —  J  =ra  log.  a  +  w  log.  l—p  log.  c. 

Ex.  5.  log.  \/a6=i(log.  a+log.  b). 

Ex.  6.  log.  y  •  ^-=£{log.  a  +  2  log.  Z>+4  log.  c-5  log.  d}. 

Ex.  7.  log.  (a3  \/^)  =  log.  (a^)  =^  log.  a, 
Ex.  8.  Iog.(a2-x2)^log.{((z+x)(a-cc)}  =log. 
Ex.  9.  log.  Va2  —  xz=  i  log.  (a+£c)+  J  log.  (a 


Ex.  10.  log.  r=i  %•  8+*  lo&  4-i  loS-  6~T  log.  2. 

6  x  v/2 


LOGARITHMS.  287 

401.  In  all  systems  of  logarithms,  the  logarithm  of  unity  is  zero. 
For  in  the  equation  ax—n, 

if  we  make  n  =  l,  the  corresponding  value  of  x  will  be  0,  since 
a°=  1,  Art.  75  ;  that  is,         log.  1  =  0. 

402.  In  all  systems  of  logarithms,  the  logarithm  of  the  base  is 
unity. 

For  a*=a; 

that  is,  log.  a  =  l. 

Common  Logarithms. 

403.  Since  the  base  of  the  common  system  of  logarithms  is 
10,  all  numbers  in  this  system  are  to  be  regarded  as  powers  of 
10.     Thus,  since 

10°=1,       we  have       log.  1  =  0; 

10!  =  10,  "  log.  10  =  1; 

102=100,         "         log.  100  =  2; 

103=1000,       "      log.  1000  =  3,  etc. 

From  this  it  appears  that  in  Briggs's  system  the  logarithm 
of  any  number  between  1  and  10  is  some  number  between  0 
and  1 ;  that  is,  it  is  a  fraction  less  than  unity,  and  is  generally 
expressed  as  a  decimal.  The  logarithm  of  any  number  between 
10  and  100  is  some  number  between  1  and  2 ;  that  is,  it  is  equal 
to  1  plus  a  decimal.  The  logarithm  of  any  number  between 
100  and  1000  is  some  number  between  2  and  3 ;  that  is,  it  is 
equal  to  2  plus  a  decimal ;  and  so  on. 

404.  The  same  principle  may  be  extended  to  fractions  by 
means  of  negative  exponents.     Thus,  since 

10-^-A-        or  0.1,       we  have       log.  0.1  = -1; 

10-^yio-      or  0.01,  "  log.  0.01  =  -2; 

10-3=T<nnr    or  0.001,         "         log.  0.001  = -3  ; 

10-4=Toiin7  or  0.0001,       "      log.  0.0001  =  -4,  etc. 
Hence  it  appears  that  the  logarithm  of  every  number  be 
tween  1  and  0.1  is  some  number  between  0  and  —1,  or  may  be 
represented  by  —1  plus  a  decimal.     The  logarithm  of  every 
number  between  0.1  and  0.01  is  some  number  between  —1  and 


288  ALGEBRA. 

—  2,  or  may  be  represented  by  —2  plus  a  decimal.  The  loga 
rithm  of  every  number  between  0.01  and  0.001  is  some  number 
between  —2  and  —3,  or  may  be  represented  by  —3  plus  a 
decimal,  and  so  on. 

405.  Hence  we  see  that  the  logarithms  of  most  numbers  must 
consist  of  two  parts,  an  integral  part  and  a  decimal  part.     The 
former  part  is  called  the  characteristic  or  index  of  the  logarithm. 
The  characteristic  may  always  be  determined  by  the  following 

RULE. 

The  characteristic  of  the  logarithm  of  any  number  is  equal  to 
the  number  of  places  by  which  the  first  significant  figure  of  that 
number  is  removed  from  the  unifs  place,  and  is  positive  when  this 
figure  is  to  the  left,  negative  ivhen  it  is  to  the  right,  and  zero  when 
it  is  in  the  unit's  place. 

Thus  the  characteristic  of  the  logarithm  of  397  is  +2,  and 
that  of  5673  is  +3,  while  the  characteristic  of  the  logarithm 
of  0.0046  is  -3. 

406.  The  same  decimal  part  is  common  to  the  logarithms  of 
all  numbers  composed  of  the  same  significant  figures. 

For,  since  the  logarithm  of  10  is  1,  it  follows  from  Art.  397 
that  if  a  number  be  divided  by  10,  its  logarithm  will  be  dimin 
ished  by  1,  the  decimal  part  remaining  unchanged.  Thus,  if 
we  denote  the  decimal  part  of  the  logarithm  of  3456  by  m,  we 
shall  have 


log.  3456  =3  + m. 
log.  345.6  =  2  +  m. 
log.  34.56  =  1  +  m. 
log.  3.456  =  0+w. 


log.  .3456=—  1+m. 
log.  .03456=  -2  +  m. 
log.  .003456  =-3  +  m. 
log.  .0003456=- 4+ m. 


Table  of  Logarithms. 

40?.  The  table  on  pages  290,  291,  contains  the  decimal  part 
of  the  common  logarithm  of  the  series  of  natural  numbers  from 
100  to  999,  carried  to  four  decimal  places.  Since  these  num 
bers  are  all  decimals,  the  decimal  point  is  omitted,  and  the  char 
acteristic  is  to  be  supplied  according  to  the  rule  in  Art.  405, 


LOGARITHMS.  289 

408.  To  find  the  logarithm  of  any  number  consisting  of  not 
more  than  three  figures. — Look  on  one  of  the  pages  of  the  ta 
ble,  along  the  left-hand  column  marked  No.,  for  the  two  left- 
hand  figures,  and  the  third  figure  at  the  head  of  one  of  the 
other  columns.     Opposite  to  the  first  two  figures,  and  in  the 
column  under  the  third  figure,  will  be  found  the  decimal  part 
of  its  logarithm.     To  this  must  be  prefixed  the  characteristic, 
according  to  the  rule  in  Art.  405.     Thus 

the  logarithm  of  347  is  2.5403; 
871  is  2.9400. 

The  logarithm  of  63,  or  63.0,  is  1.7993; 
"  5,  or  5.00,  is  0.6900; 

"  0.235  is  1.3711. 

The  minus  sign  is  here  placed  over  the  characteristic,  to  show 
that  that  alone  is  negative,  while  the  decimal  part  of  the  loga 
rithm  is  positive. 

409.  To  find  the  logarithm  of  any  number  containing  more 
than  three  figures. — By  inspecting  the  table,  we  shall  find  that 
within  certain  limits  the  differences  of  logarithms  are  propor 
tional  to  the  differences  of  their  corresponding  numbers.     Thus 

the  logarithm  of  216  is  2.3345; 
"  217  is  2.3365; 

"  218  is  2.3385. 

Here  the  difference  between  the  successive  logarithms,  called 
the  tabular  difference,  is  constantly  20,  corresponding  to  a  differ 
ence  of  unity  in  the  natural  numbers.  If,  then,  we  suppose  the 
logarithms  to  increase  at  the  same  rate  as  their  corresponding 
numbers  (as  they  do  nearly),  a  difference  of  0.1  in  the  numbers 
should  correspond  to  a  difference  of  2  in  the  logarithms;  a  dif 
ference  of  0.2  in  the  numbers  should  correspond  to  a  differ 
ence  of  4  in  the  logarithms,  etc.  Hence 

the  logarithm  of  216.1  must  be  2.3347; 

216.2        "        2.3349,  etc. 

In  order  to  facilitate  the  computation,  there  is  given,  on  the 
right  margin  of  each  page,  the  proportional  part  for  the  fourth 
figure  of  the  natural  number,  corresponding  to  tabular  differ- 

N 


290 


ABLE   OF   COMMON   LOGARITHMS. 


•'NO. 

10 
1  1 

12 

i3 
i4 
i5 
16 

17 

18 

X9 

20 
21 
22 
23 
24 
25 
26 
27 

0 

i 

2 

3 

4  |  5 

6 

7 

8 

9 

PE 

I 
2 
3 

4 
5 
6 

7 

8 

9 

i 

2 

3 
4 
5 
6 

7 
8 

9 
i 

2 

3 
4 
5 
6 

8 
9 

i 

2 

3 

4 
5 
6 

8 
9 

OPO 

43 
~ 

17 

22 
26 

3o 
34 
39 

38 

~4 
8 
ii 
i5 
J9 

23 

27 
3o 
34 

33 

RTIO 

42 

T 

8 
i3 
'7 

21 
25 
29 

34 
38 

_37_ 

4 

7 
ii 
i5 
19 

22 
26 

3o 
33 

32 

*ML 

4i 

PA] 

4o 

ITS 

3( 

oooo 
o4i4 
0792 
1  1  39 
i46i 
1761 

2o4l 

23o4 
2553 

oo43 
o453 
0828 
n73 
1492 
1790 
2068 
233o 
2577 

0086 
0492 

o864 
1206 
i523 
1818 
2og5 
2355 
2601 

0128 
o53i 
o899 

I239 

i553 

i847 

2122 
2380 
2625 

OI7O 

o569 
o934 

I27I 

1  584 
i875 
2148 
24o5 
2648 

0212 

o6o7 
0969 
i3o3 
i6i4 
1903 

2I75 
2430 
2672 

0253 
o645 
ioo4 
i335 
1  644 
i93i 

2201 

2455 
2695 

O294 
0682 

io38 
i367 
i673 
i959 

2227 
2480 
27l8 

o334 
o7i9 

IO72 

i399 
I7o3 
i987 

2253 

25o4 

2742 

o374 
o755 
1  106 
i43o 

I732 
20l4 
2279 

2529 

2765 

4 

8 

12 

z6 

21 
25 
29 

33 

37 

36 

"7 

7 
1  1 

i4 
18 

22 
25 
29 
32 

3i 

4 
8 

12 

16 

20 

24 

28 

36 

35 

~ 

7 
1  1 

i4 

18 

21 

25 

28 

32 

3o 

i 

\ 
i: 
ij 

2C 
2! 
2r 

3: 

3J 

3z 
K 

u 

i- 
24 

2/ 

3; 

24 

2788 
3oio 

3222 

3424 

36i7 

38o2 
3979 
4i5o 
43:4 

2810 
3o32 
3243 
3444 
3636 
3820 
3997 
4i66 
433o 

2833 
3o54 
3263 
3464 
3655 
3838 
4oi4 
4i83 
4346 

2856 
3o75 
3284 
3483 
3674 
3856 
4o3i 
4200 
4362 

2878 
3096 
33o4 
35o2 
3692 
3874 
4o48 
4216 
4378 

2900 

3n8 
3324 

3522 

37n 

3892 
4o65 

4232 

4393 

2923 

SiSg 

3345 
354i 
3729 
39o9 
4082 
4249 
44o9 

2945 
3  1  60 
3365 
356o 
3747 
3927 
4o99 
4265 
44s5 

2967 

3i8i 
3385 
3579 
3766 
3945 
4n6 
4281 
444o 

2989 

3201 

34o4 
3598 
3784 
3962 
4i33 
4298 
4456 

28 
29 

3o 
3i 

32 

33 
34 
35 
36 

37 
38 
39 
4o 

4i 

42 

43 
44 
45 

46" 
47 
48 

r49 
5o 
5i 

5  2 

53 

54 

4472 
4624 
4771 
4914 
5o5i 
5i85 
53i5 
544  1 
5563 

4487 
4639 
4786 
4928 
5o65 
5i98 
5328 
5453 
5575 

45o2 
4654 
48oo 
4g42 
5o79 

521  I 

534o 

5465 
5587 

45i8 
4669 
48i4 
4955 
5092 

5224 

5353 
5478 
5599 

4533 
4683 
4829 
4969 
5io5 
5237 
5366 
5490 
56n 

4548 
4698 
4843 
4983 
5i  19 
525o 
5378 
55o2 
5623 

4564 
47i3 

4857 

4997 
5  1  32 
5263 
5391 
55i4 
5635 

4579 
4728 
487i 
5oi  i 
5i45 
5276 
54o3 
5527 
5647 

4594 
4742 
4886 
5024 
5i59 
5289 
54i6 
5539 
5658 

46o9 

4?5? 
49oo 

5o38 
5172 
53o2 
5428 
555i 
5670 

3 

7 
10 

i3 

J7 
20 

23 

26 
3o 

28 

3 
6 
10 
i3 
16 
*9 

22 
26 
29 

27 

3 
6 
9 

12 

16 
19 

22 

25 

28 
26 

3 
6 
9 

12 

i5 

18 

21 

24 
27 

25 

( 
( 
i: 
I< 
!• 

2( 
2v 
2( 

•2i 

5682 
5798 
59n 
6021 
6128 
6232 
6335 
6435 
6532 

5694 
58o9 
5922 
6o3i 
6i38 
6243 
6345 
6444 
6542 

57o5 
582i 
5933 
6042 
6i49 
6253 
6355 
6454 
655i 

57i7 
5832 
5944 
6o53 
6160 
6263" 
6365 
6464 
656i 

5729 
5843 
5955 
6064 
6i7o 
6274 
6375 
64  74 
65  7  1 

574o 
5855 
5966 
6o75 
6180 
6284 
6385 
6484 
658o 

5752 
5866 
5977 
6o85 
6191 
6294 
6395 
6493 
659o 

5763 
5877 
5988 
6096 
6201 
63o4 
64o5 
65o3 
6599 

5775 

5888 

5999 
6io7 

6212 
63i4 
64i5 
65i3 
6609 

5786 
5899 
6010 
6117 
6222 
6325 
6425 

6522 

6618 

3 
6 
8 
ii 
i4 

*7 

20 

22 
25 

3 
5 
8 
1  1 
i4 
16 

*9 

24 

3 
5 
8 

10 

i3 
16 
18 

21 

23 

3 
5 

8 

10 

i3 

i5 
1  8* 

20 
23, 

( 

1C 
It 

\L 

l"t 

<5 

H 

6628 
6721 
6812 

6902 

699o 
7076 
7160 
7243 
7324 

6637 
673o 
6821 
691  1 
6998 
7o84 
7i68 
y25  1 
7332 

6646 
6739 
683o 
6920 
7oo7 

7°93 

7177 
7259 
734o 

6656 
6749 
68  °9 
6928 
7oi6 

7IOI 

7i85 

7267 

7348 

6665 
6758 
6848 
6937 

7024 
7I  10 

7I93 

7275 

7356 

6675 
6767 
68  5  7 
6946 
7o33 
7118 
7202 
7284 
7364 

6684 
6776 
6866 
6955 
7042 
7126 
7210 

7292 
7372 

6693 
6785 
6875 
6964 
7o5o 
7i35 

72l8 

73oo 
738o 

6702 

6794 
6884 
69-72 
7o59 
7i43 

7226 

73o8 
7388 

6712 
68o3 
6893 
698i 
7067 
7i52 
7235 
73i6 
7396 

TABLE  OF  COMMON  LOGARITHMS. 


No. 

o 

i 

2 

3 

4 

5 

6 

7 

8 

9 

PR 

I 

2 

3 

4 
5 
6 

8 
9 

i 

2 

3 

4 
5 
6 

8 
9 

i 

2 

3 
4 
5 
6 

8 
9 

i 

2 

3 

4 
5 
6 

7 
8 

Q 

OPOl 

23 
2 

5 
7 
9 

i4 
16 
18 

21 

18 

2 

4 
5 

7 

9 
ii 
i3 
i4 
16 

i3 

mo 

22 
2 

4 
7 
9 
1  1 
i3 
i5 
18 
20 

'7 

2 

3 
5 
7 
9 
10 

12 
14 

i5 

12 

NAL 

21 

2 

4 
6 
8 
1  1 
i3 
i5 
i? 
19 

2L 

2 

3 
5 
6 

8 

10 

ii 
i3 
i4 

ii 

PAI 

20 
2 

4 
0 
8 
10 

12 

i4 
16 

18 

2L 

2 

3 
5 
6 
8 

9 
ii 

12 

i4 

10 

tTS 

12 

2 
i 

6 
8 
10 
ii 
i3 
i5 
17 

2* 

i 

3 
4 
6 

8 
10 
1  1 
i3 

9 

55 
56 

57 

58 
59 
60 
61 
62 
63 

?4o4 
7482 
7559 
7634 
7709 
7782 
7853 
7924 
7993 

74l2 

749o 
7566 
•7642 
77i6 
7789 
7860 
793i 
8000 

74i9 
7497 
7574 
764g 
7723 

7796 

7868 

7938 
8oo7 

7427 

75o5 
7582 
7657 
773i 
78o3 
7875 
7945 
8oi4 

7435 
75i3 
7589 
7664 
7738 
78io 
7882 

7952 

8021 

7443 

7520 

7597 

7672 

7745 
78i8 
7889 
7959 
8028 

745i 
7528 
76o4 
7679 
7752 
7825 
7896 
7966 
8o35 

7459 
7536 

76l2 

7686 
776o 
7832 
79°3 
7973 
8o4i 

7466 
7543 

7619 
7694 
7767 
7839 
79io 

798° 
8o48 

7474 
755i 
7627 
7701 
7774 
7846 
79i7 

7987 
8o55 

64 
65 
66 
67 
68 
69 
7° 
71 
72 

73 

74 
75 
76 

77 

78 

79 
80 
81 

"sT 

83 
84 
85 
86 
87 
'88 

•89 

J?l 

91 
92 
93 

94 
95 
96 

97 

98 

99 

8062 
8129 
8i95 
8261 
8325 
8388 
845  1 
85i3 
8573 

"8633" 
8692 

875i 
8808 
8865 

892I 

8976 
9o3i 
9o85 

8o69 
8i36 
8202 
8267 
833i 
8395 
8457 
85i9 
8579 

8o75 
8142 
8209 
8274 
8338 
84oi 
8463 
8525 
8585 

8082 
8i49 
82i5 
8280 
8344 
84o7 
847o 
853i 
859i 

8o89 
8i56 
8222 
8287 
835i 
84i4 
8476 
8537 
8597 

8o96 
8162 
8228 
8293 
8357 
8420 
8482 
8543 
86o3 

8102 
8i69 
8235 
8299 
8363 
8426 
8488 
8549 
86o9 

8io9 
8176 
8241 
83o6 
8370 
8432 
8494 
8555 
86i5 

8116 
8182 
8248 
83i2 
8376 
8439 
85oo 
856i 
8621 

8122 

8i89 
8254 
83i9 
8382 
8445 
85o6 
8567 
8627 

8639 
8698 
8756 
88i4 
887i 
8927 
8982 
9o36 
9o9o 

8645 
87o4 
8762 
8820 
8876 
8932 
8987 
9042 
9096 

865i 
87io 

8768 
8825 
8882 
8938 
8993 
9o47 

9IOI 

8657 
87i6 
8774 
883i 
8887 
8943 
8998 
9o53 
9io6 

8663 
8722 

8779 
8837 
8893 
8949 
9oo4 
9o58 

9I  12 

8669 
8727 
8785 
8842 
8899 
8954 
9oo9 
9o63 
9n7 

8675 
8733 
879i 
8848 
89o4 
896o 
9oi5 
9o69 

9I22 

8681 
8739 

8797 
8854 
89io 
8965 

9020 

9o74 

9I28 

8686 
8745 
8802 
8859 
89i5 
897i 

9025 

9°79 
9i33 

i 
3 
4 
5 

7 

8 

9 

10 
12 

8 

I 

2 

4 
5 
6 

7 

8 

10 

ii 

7 

i 

2 

3 
4 
6 

7 

8 

9 

10 

C 

I 
2 

3 
4 
5 
6 

7 
8 

9 

5 

i 

2 

3 
4 
5 
5 
6 

8 

4 

o 
i 
i 

2 
2 

2 

3 
3 

4 

9i38 
9i9i 
9243 
9294 
9345 
9395 
9445 
9494 
9542 

9i43 
9i96 

9248 

9299 
935o 
94oo 
945o 
9499 
9547 

9149 
9201 
9253 
93o4 
9355 
94o5 
9455 
95o4 
9552 

9i54 
92o6 
9258 

9^9 
936o 

94io 
946o 

9<?9 
9557 

9i59 

92I2 
9263 

93i5 
9365 
94i5 
9465 
95i3 
9562 

9i65 

92I7 

9269 

9320 

937o 

9420 

9469 
95i8 
9566 

9i7o 

9222 

9274 

9325 
9375 
94s5 
9474 
9533 
957i 

9i75 

9227 
9279 

933o 
938o 
943o 

9479 
9528 
9576 

9i8o 

9232 

9284 

9335 
9385 
9435 
9484 
9533 
958i 

9i86 
9238 
9289 
934o 
939o 
944o 
9489 
9538 
9586 

i 

2 
2 
3 

4 
5 
6 
6 
7 

i 
i 

2 

3 
4 
4 

5 
6 
6 

i 
i 

2 
2 

? 

4 
4 
5 
5 

i 
i 

2 

2 

3 
3 
4 
4 
S 

959o 
9638 
9685 
973i 

9777 
9823 
9868 
9912 
9956 

9595 
9643 
9689 
9736 
9782 
9827 
9872 
9917 
9961 

96oo 
9647 
9694 
974i 
9786 
9832 
9877 

992I 

9965 

96o5 
9652 

9699 
9745 
979i 
9836 
988i 
9926 
9969 

96o9 
9657 
9703 
975o 
9795 
984i 
9886 
993o 
9974 

96  1/' 
966i 
97o8 
9754 
98oo 
9845 
989o 
9934 
9978 

96i9 
9666 
97i3 
9759 
98o5 
985o 
9894 
9939 
9983 

9624 
967i 
97i7 
9763 
98o9 
9854 

9899 
9943 

9987 

9628 
9675 

9722 

9768 
98i4 
9859 
99o3 
9948 
999  ' 

9633 
968o 

9727 
9773 
98i8 
9863 
99o8 
9952 
99q6 

292  ALGEBRA. 

ences  from  43  to  4.  Thus,  on  page  291,  near  the  top,  we  see 
that  when  the  tabular  difference  is  20,  the  corrections  for  .1, 
.2,  .3,  etc.,  are  2,  4,  6,  etc. 

It  is  obvious  that  the  correction  for  a  figure  in  the  fifth  place 
of  the  natural  number  must  be  one  tenth  of  the  correction  for 
the  same  figure  if  it  stood  in  the  fourth  place.  Such  a  correc 
tion  would,  however,  generally  be  inappreciable  in  logarithms 
which  extend  only  to  four  decimal  places. 

EXAMPLES. 

Find  the  logarithm  of  4576.  Ans.  3.6605. 

13.78.  Ans.  1.1392. 

"  "  1.682.  Ans.  0.2258. 

.03211.  4ws._2.5066. 

"  "  .4735.  Ans.  1.6753. 

15983.  Ans.  4.2036. 

The  logarithms  here  given  are  only  approximate.  We  can 
obtain  the  exact  logarithm  of  very  few  numbers ;  but  by  taking 
a  sufficient  number  of  decimals  we  can  approach  as  nearly  as 
we  please  to  the  true  logarithm. 

410.  To  find  the  natural  number  corresponding  to  any  loga 
rithm. — Look  in  the  table  for  the  decimal  part  of  the  loga 
rithm,  neglecting  the  characteristic;  and  if  the  decimal  is  ex 
actly  found,  the  first  two  figures  of  the  corresponding  natural 
number  will  be  found  opposite  to  it  in  the  column  headed  No., 
and  the  third  figure  will  be  found  at  the  top  of  the  page.  This 
number  must  be  made  to  correspond  with  the  characteristic  by 
pointing  off  decimals  or  annexing  ciphers.  Thus 
the  natural  number  belonging  to  the  logarithm  3.3692  is  2340; 
"  "  "  "  1.5378  is  34.5. 

If  the  decimal  part  of  the  logarithm  is  not  exactly  contained 
in  the  table,  look  for  the  nearest  less  logarithm,  and  take  out 
the  three  figures  of  the  corresponding  natural  number  as  be 
fore.  The  additional  figure  or  figures  may  be  obtained  by 
means  of  the  proportional  parts  on  the  margin  of  the  pnge. 

Find  the  number  corresponding  to  the  logarithm  3.3685. 


LOGARITHMS.  293 

The  next  less  logarithm  in  the  table  is  .3674,  and  the  three 
corresponding  figures  of  the  natural  number  are  233.  Their 
logarithm  is  less  than  the  one  proposed  by  11,  and  the  tabular 
difference  is  18.  By  referring  to  the  margin  of  page  291,  we 
find  that,  with  a  difference  of  18,  the  figure  corresponding  to 
the  proportional  part  11  is  6.  Hence,  since  the  characteristic 
of  the  proposed  logarithm  is  3,  the  required  natural  number  is 
'2336. 

EXAMPLES. 

1.  Find  the  number  corresponding  to  the  logarithm  2.5386. 

Ans.  345.6. 

2.  Find  the  number  corresponding  to  the  logarithm  0.2345. 

Ans.  1.716. 

3.  Find  the  number  corresponding  to  the  logarithm  1.9946. 

Ans.  9^.76. 

4.  Find  the  number  corresponding  to  the  logarithm  1.6478. 

Ans.  0.4444. 

411.  Multiplication  ~by  Logarithms. — According  to  Art.  396, 
to  find  the  product  of  two  numbers  we  have  the  following 

RULE. 

Add  the  logarithms  of  the  factors  ;  the  sum  will  be  the  logarithm 
of  the  product. 

The  word  sum  is  here  to  be  understood  in  its  algebraic  sense. 
The  decimal  part  of  a  logarithm  is  invariably  positive ;  but  the 
characteristic  may  be  either  positive  or  negative. 

Ex.  1.  Find  the  product  of  57.98  by  3.12. 

The  logarithm  of  57.98  is  1.7633. 

3.12  is  0.4942. 

The  log.  of  the  product  180.9  is  2.2575. 

Ex.  2.  Find  the  product  of  0.00563  by  172.5. 

The  logarithm  of         0.00563  is  "3.7505. 
"  172.5  is  2.2368. 

The  log.  of  the  product  0.971  is  L9873. 

Ex.  3.  Find  the  product  of  54.32  by  6.543. 

Ex.  4.  Find  the  product  of  3.854  by  0.5761. 


294  ALGEBRA. 

412.  Division  by  Logarithms. — According  to  Art.  397,  to  find 
the  quotient  of  two  numbers  we  have  the  following 

KULE. 

From  the  logarithm  of  the  dividend  subtract  the  logarithm  of 
the  divisor ;  the  difference  will  be  the  logarithm  of  the  quotient. 

The  word  difference  is  here  to  be  understood  in  its  algebraic 
sense ;  the  decimal  part  of  the  logarithm  being  invariably  pos 
itive,  while  the  characteristic  may  be  either  positive  or  nega 
tive. 

Ex.  1.  Find  the  quotient  of  888.7  divided  by  42.24. 

The  logarithm  of  888.7  is  2.9488. 

"  42.24  is  1.6257. 

The  quotient  is  21.04,  whose  log.  is  1.3231. 

Ex.  2.  Find  the  quotient  of  0.8692  divided  by  42.32. 

The  logarithm  of  0.8692  is  T.9391. 

"  42.32  is  1.6265. 

The  quotient  is    0.02054,  whose  log.  is  2^3126. 

Ex.  3.  Find  the  quotient  of  380.7  divided  by  13.75. 
Ex.  4.  Find  the  quotient  of  24.93  divided  by  .0785. 

413.  Involution  by  Logarithms.  —  According  to  Art.  398,  to 
involve  a  number  to  any  power  we  have  the  following 

RULE. 

Multiply  the  logarithm  of  the  number  by  the  exponent  of  the 
power  required. 

It  should  be  remembered  that  what  is  carried  from  the  deci 
mal  part  of  the  logarithm  is  positive,  whether  the  characteris 
tic  be  positive  or  negative. 

Ex.  1.  Find  the  fifth  power  of  2.846. 

The  logarithm  of  2.846  is  0.4542. 

5 

The  fifth  power  is  186.65,  whose  log.  is  2.2710. 

Ex.  2.  Find  the  cube  of  .07654. 


LOGARITHMS.  295 

The  logarithm  of  .07654  is  2^8839. 

3 

The  cube  is  0.0004484,  whose  log.  is  46517, 

Ex.  3.  Find  the  20th  power  of  1.06. 
Ex.  4.  Find  the  seventh  power  of  0.8952. 

414.  Evolution  by  Logarithms. — According  to  Art.  399,  to 
extract  any  root  of  a  number  we  have  the  following 

RULE. 

Divide  the  logarithm  of  the  number  by  the  index  of  tJie  root  re 
quired. 

Ex.  1.  Find  the  cube  root  of  482.4. 

The  logarithm  of  482.4  is  2.6834. 

Dividing  by  3,  we  have  0.8945,  which  corresponds  to  7.843, 
which  is  therefore  the  root  required. 

Ex.  2.  Find  the  100th  root  of  365.  Ans.  1.061. 

When  the  characteristic  of  the  logarithm  is  negative,  and  is 
not  divisible  by  the  given  divisor,  we  may  increase  the  charac 
teristic  by  any  number  which  will  make  it  exactly  divisible, 
provided  we  prefix  an  equal  positive  number  to  the  decimal 
part  of  the  logarithm. 

Ex.  3.  Find  the  seventh  root  of  0.005846. 

The  logarithm  of  0.005846  is  "3.7669,  which  may  be  written 

7+4.7669. 

Dividing  by  7,  we  have  1.6810,  which  is  the  logarithm  </ 
.4797,  which  is  therefore  the  root  required. 

Ex.  4.  Find  the  10th  root  of  0.007815. 

415.  Proportion  by  Logarithms. — The  fourth  term  of  a  pro 
portion  is  found  by  multiplying  together  the  second  and  third 
terms  and  dividing  by  the  first.     Hence,  to  find  the  fourth  term 
of  a  proportion  by  logarithms,  we  have  the  following 


296  ALGEBRA. 

RULE. 

Add  the  logarithms  of  the  second  and  third  terms,  and  from 
sum  subtract  the  logarithm  of  the  first  term. 

Ex.  1.  Find  a  fourth  proportional  to  72.34,  2.519,  and  357.5. 

Ans.  12.45. 

Ex.  2.  Find  a  fourth  proportional  to  43.17,  275,  and  5.762. 
Ex.  3.  Find  a  fourth  proportional  to  5.745,  781.2,  and  54.27. 

Exponential  Equations. 

416.  An  exponential  equation   is  one  in  which  the  unknown 
quantity  occurs  as  an  exponent.     Thus, 

ax  =  b 

is  an  exponential  equation,  from  which,  when  a  and  ~b  are 
known,  the  value  of  x  may  be  found.  If  a  — 2  and  Z>  =  8,  the 
equation  becomes  2^  —  8, 

in  which  the  value  of  x  is  evidently  3,  since  23  — 8. 
If  a  =  16  and  Z>  =  2,  the  equation  becomes 

16*=2, 

in  which  the  value  of  x  is  evidently  J,  since  16—2. 

417.  Solution  by  Logarithms. — When  b  is  not  an  exact  power 
or  root  of  a,  the  equation  is  most  readily  solved  by  means  of 
logarithms.      Taking  the  logarithm  of  each  member  of  the 
equation  ax  —  b,  we  have 

x  log.  a=log.  &, 

,  log.  1) 

whence  X=Y~    —. 

log.  a 

Ex.  1.  Solve  the  equation  3* =20. 

log.  20     1.3010 

x  =  -r-   — r~    A^~,  =2.727  nearly, 
log.  3       .47v  1 

Ex.  2.  Solve  the  equation  5*  =  12. 

(2V 
-j  =J. 

Ex.  4.  Solve  the  equation  10* =7. 

i 

Ex.  5.  Solve  the  equation  12* =3. 

3 

Ex.  6.  Solve  the  equation  12*  =  To 


LOGARITHMS.  297 

Compound  Interest. 

418.  Interest  is  money  paid  for  the  use  of  money.     When  the 
interest,  as  soon  as  it  becomes  due,  is  added  to  the  principal,  and 
interest  is  charged  upon  the  whole,  it  is  called  compound  interest. 

419.  To  find  the  amount  of  a  given  sum  in  any  time  at  com 
pound  interest.     It  is  evident  that  $1.00  at  5  per  cent,  interest 
becomes  at  the  end  of  the  year  a  principal  of  $1.05  ;  and,  since 
the  amount  at  the  end  of  each  year  must  be  proportioned  to 
the  principal  at  the  beginning  of  the  year,  the  amount  at  the 
end  of  two  years  will  be  given  by  the  proportion 

1.00: 1.05::  1.05 :  (1.05)2. 

The  sum  (1.05)2  must  now  be  considered  as  the  principal, 
and  the  amount  at  the  end  of  three  years  will  be  given  by  the 
proportion 

1.00: 1.05  ::(1.05)2:(1.05)3. 

In  the  same  manner,  we  find  that  the  amount  of  $1.00  for 
n  years  at  5  per  cent,  compound  interest  is  (1.05)n. 

For  the  same  reason,  the  amount  for  n  years  at  6  per  cent,  is 
(1.06)n.     It  is  also  evident  that  the  amount  of  P  dollars  for  a 
given  time  must  be  P  times  the  amount  of  one  dollar. 
Hence,  if  we  put 

P  to  represent  the  principal, 
r  the  interest  of  one  dollar  for  one  year, 
n  the  number  of  years  for  which  interest  is  taken, 
A  the  amount  of  the  given  principal  for  n  years, 
we  shall  have  A  =  P  (1 + r)n. 

This  equation  contains  four  quantities,  A,  P,  n,  r,  any  three 
of  which  being  given,  the  fourth  may  be  found.  The  compu 
tation  is  most  readily  performed  by  means  of  logarithms.  Tak 
ing  the  logarithms  of  both  members  of  the  preceding  equation 
and  reducing,  we  find 

log.  A=log.  P  +  nxlog.  (1  +  r), 
log.  P  =  log.  A  —  n  X  log.  ( 1  -f  r), 

,       /-,   ,    N     log.  A  — log.  P 
log.(l+r)=-         n     -     , 

_log.  A -log.  P 


298  ALGEBRA. 

Ex.  1.  How  much  would  500  dollars  amount  to  in  five  years 
at  6  per  cent,  compound  interest  ? 

The  log.  of  1.06  is  0.0253 

6 

0.1265 

The  log.  of  500  is  2.6990 

The  amount  is  $669.10,  whose  log.  is  2.8255. 

Ex.  2.  What  principal  at  6  per  cent,  compound  interest  will 
amount  to  500  dollars  in  seven  years?  Ans.  $332.60. 

Ex.  3.  At  what  rate  per  cent,  must  500  dollars  be  put  out  at 
compound  interest  so  that  it  may  amount  to  $680.30  in  seven 
years  ?  Ans.  4%  per  cent. 

Ex.  4.  In  what  time  will  500  dollars  amount  to  900  dollars 
at  6  per  cent,  compound  interest  ?  Ans.  10TTT  years. 

Ex.  5.  How  much  would  400  dollars  amount  to  in  nine  years 
at  5  per  cent,  compound  interest? 

Ex.  6.  What  principal  at  5  per  cent,  compound  interest  will 
amount  to  400  dollars  in  eight  years? 

Ex.  7.  At  what  rate  per  cent,  must  400  dollars  be  put  out 
at  compound  interest  so  that  it  may  amount  to  $620.70  in  nine 
years  ? 

Ex.  8.  In  what  time  will  a  sum  of  money  double  at  6  per 
cent,  compound  interest? 

Ex.  9.  In  what  time  will  a  sum  of  money  double  at  5  per 
cent,  compound  interest? 

Annuities. 

420.  An  annuity  is  a  sum  of  money  stipulated  to  be  paid  an 
nually,  and  to  continue  for  a  given  number  of  years,  for  life, 
or  forever. 

421.  To  find  the  amount  of  an  annuity  left  unpaid  for  any  num* 
her  of  years,  allowing  compound  interest. 

Let  a  denote  the  annuity,  n  the  number  of -years,  r  the  in 
terest  of  one  dollar  for  one  year,  and  A  the  required  amount. 
The  amount  due  at  the  end  of  the  first  year  is  a. 
At  the  end  of  the  second  year  the  amount  of  the  first  an- 


LOGARITHMS.  299 

unity  is  a(l-f  r),  and  a  second  payment  becomes  due;  hence 
the  whole  sum  due  at  the  end  of  the  second  year  is  a  +  a(l-f  r). 

At  the  end  of  the  third  year  a  third  payment  a  becomes  due, 
together  with  the  interest  on  a  -{-a  (I  -\-r)  ;  hence  the  whole  sum 
due  at  the  end  of  the  third  year  is  a+a(l-|-r)-|-a(l-f  r)2,  or 
a{l  +  (l  +  r)-f  (1  +  r)2},  and  so  on. 

Hence  the  amount  due  at  the  end  of  n  years  is 


These  terms  form  a  geometrical  progression  in  which  the  ratio 
is  1+r.     Hence,  by  Art.  332,  the  sum  of  the  series  is 


422.  To  find  the  present  value  of  an  annuity,  to  continue  for  a 
certain  number  of  years,  allowing  compound  interest. 

The  present  value  of  the  annuity  must  be  such  a  sum  as, 
if  put  out  to  interest  for  n  years  at  the  rate  r,  would  amount 
to  the  same  as  the  amount  of  the  annuity  at  the  end  of  that 
period. 

If  P  denote  the  present  value  of  the  annuity,  then  the  amount 
of  the  annuity  will  be  P(l-fr)n,  which  must  be  equal  to 

a. 


Therefore  p=g-(l+r)'-    1 


r 

Ex.  1.  How  much  will  an  annuity  of  500  dollars  amount  to 
in  15  years  at  four  per  cent,  compound  interest  ? 

(l+r)»       =1.7987 

(l+r)n—  1  =  .7987,  whose  log.  is  L9024 

the  log.  of  .04  is  2".6Q21 

L3003 

the  log.  of  500  is  2.6990 
The  amount  is  $9983,  whose  log.  is  3.9993. 

Ex.  2.  What  is  the  present  value  of  an  annuity  of  500  dol- 


300  ALGEBKA. 

lars  to  continue  for  20  years,  interest  being  allowed  at  the  rate 
of  four  per  cent,  per  annum  ? 

(l+r)n       =2.188 

(l;+r)*— 1=1.188,  whose  log.  is  0.0748 
the  log.  of  (l+r)n  is  0.3400 

1.7348 
-=12500,  whose  log.  is  4.0969 


The  present  value  is  $6787,  whose  log.  is  3.8317. 

Ex.  3.  How  much  will  an  annuity  of  600  dollars  amount  to 
in  12  years  at  three  per  cent,  compound  interest? 

Ex.  4.  What  is  the  present  value  of  an  annuity  of  600  dollars 
to  continue  for  12  years  at  three  per  cent,  compound  interest  ? 

Ex.  5.  In  what  time  will  an  annuity  of  500  dollars  amount 
to  5000  dollars  at  4  per  cent,  compound  interest? 

Ans.  In  8-f  years, 

Increase  of  Population. 

423.  The  natural  increase  of  population  in  a  country  is  some 
times  computed  in  the  same  way  as  compound  interest.  Know 
ing  the  population  at  two  different  dates,  we  compute  the  rate 
of  increase  by  Art.  419,  and  from  this  we  may  compute  the  pop 
ulation  at  any  future  time  on  the  supposition  of  a  uniform  rate 
of  increase.  Such  computations,  however,  are  not  very  reliable, 
for  in  some  countries  the  population  is  stationary,  and  in  others 
it  is  decreasing. 

Ex.  1.  The  number  of  the  inhabitants  of  the  United  States 
in  1790  was  3,930,000,  and  in  1860  it  was  31,445,000.  What 
was  the  average  increase  for  every  ten  years  ? 

Ans.  34§  per  cent. 

Ex.  2.  Suppose  the  rate  of  increase  to  remain  the  same  for 
the  next  ten  years,  what  would  be  the  number  of  inhabitants 
in  1870?  Ans.  42,330,000. 

Ex.  3.  At  the  same  rate,  in  what  time  would  the  number  in 
1860  be  doubled  ?  Ans.  23£  years. 

Ex.  4.  At  the  same  rate,  in  what  time  would  the  number  in 
1860  be  tripled? 


LOGARITHMS.  301 

To  find  the  Logarithm  of  any  given  Number. 
424.  If  m  and  n  denote  any  two  numbers,  and  x  and  y  their 

/y»     1      si  i 

logarithms,  then  — — —  will  be  the  logarithm  of  Vmn.     For,  ac 
cording  to  Art.  396,  ax+y—mn,  and,  taking  the  square  root  of 

x+y  . 

each  member,  we  have  a  2  —Vmn.     Therefore,         ^  is  the 

logarithm  of  Vmn,  since  it  is  the  exponent  of  that  power  of 
the  base  which  is  equal  to  Vmn. 

Now,  in  Briggs's  system,  the  logarithm  of  10  is  1,  of  100  is 

14-2 
2,  etc.     Hence  the  logarithm  of  VlO  x  100  is  ~  -  ;  that  is,  the 

Zi 

logarithm  of  31.6228  is  1.5. 


1  +  1.5 


So,  also,  the  logarithm  of  VlO  x  31.6228  is  -       - ;  that  is, 

2 

the  logarithm  of  17.7828  is  1.25,  and  so  on  for  any  number  of 
logarithms. 

In  this  manner  were  the  first  logarithmic  tables  computed ; 
but  more  expeditious  methods  have  since  been  discovered.  It 
is  found  more  convenient  to  express  the  logarithm  of  a  number 
in  the  form  of  a  series. 

425.  Logarithms  computed  ly  Series. — The  computation  of  log 
arithms  by  series  requires  the  solution  of  the  equation 


in  which  a  is  the  base  of  the  system,  n  any  number,  and  x  is 
the  logarithm  of  that  number.  In  order  that  a  and  n  may  be 
expanded  into  a  series  by  the  binomial  theorem,  we  will  con 
vert  them  into  binomials,  and  assume  a  =  l-\-b  and  n  =  l  +  m; 
then  we  shall  have 


where  x  is  the  logarithm  of  1  +  m,  to  the  base  1-f  &,  or  a. 
Involving  each  member  to  a  power  denoted  by  ?/,  we  have 


Expanding  both  members  by  the  binomial  theorem,  we  have 


302  ALGEBKA. 

xy(xy—  1)79     xy(xy— 


i  eto. 


.  o 

Canceling  unity  from  both  members  and  dividing  by  y,  we 
have 

t  eto.)= 


i  etc. 


This  equation  is  true  for  all  values  of  y  ;  it  will  therefore  be 
true  when  y=0.     Upon  this  supposition,  the  equation  becomes 

,,     b2     b3         .    .  w2     m3 

35  (&—  g  +-3-'  etc.)=m  —  —  +-  —  ,  etc., 

m2     m3 
m-  —  +  -g  —  ,etc. 

whence  x=log.  (l+m)=  --  -  —  —       —  . 


If  we  put  M= 


the  last  equation  becomes 

(yyj2  *yyi  3 

cc^log.  w-log.  (l  +  m)  =  M(m—  —  +—  —  ,  etc.).     (1.) 

'We  have  thus  obtained  an  expression  for  the  logarithm  of 
the  number  1-f-m  or  n.  This  expression  consists  of  two  fac 
tors,  viz.,  the  quantity  M,  which  is  constant,  since  it  depends 
simply  upon  the  base  of  the  system  ;  and  the  quantity  within 
the  parenthesis,  which  depends  upon  the  proposed  number. 
The  constant  factor  M  is  called  the  modulus  of  the  system. 

426.  To  determine  the  Base  of  Napier's  System.  —  In  Napier's 
system  of  logarithms  the  modulus  is  assumed  equal  to  unity. 
From  this  condition  the  base  may  be  determined.  Equation 
(1),  Art.  425,  in  this  case  becomes 


m3     m4 


x=m-—+—  -—  +,  etc. 


LOGARITHMS.  303 

Reverting  this  series,  Art.  883,  Ex.  3,  we  obtain 

X2         X3  X* 

w=a:+¥+0+0^+'  etc' 

But,  by  hypothesis,  ax  —  n  —  l  +  m;  therefore 

X2         X3  CC4 


If  x  be  taken  equal  to  unity,  we  have 


By  taking  nine  terms  of  this  series,  we  find 

a=2.718282, 
which  is  the  base  of  Napier's  system. 

427.  The  logarithm  of  a  number  in  any  system  is  equal  to  tJw 
modulus  of  that  system  multiplied  by  the  Naperian  logarithm  of  the 
number. 

If  we  designate  Naperian  logarithms  by  Nap.  log.,  and  log 
arithms  in  any  other  system  by  log.,  then,  since  the  modulus 
of  Napier's  system  is  unity,  we  have 

m?     m3 
log.  (l+m)  =  M(m-      +-,  etc.), 


Nap.  log.(l  +  7n)=m—    -+    —  ,  etc. 

/i  O 

Hence  log.  (1  +  m)  =  M  x  Nap.  log.  (1  -f  m), 

or  M 


Nap.  log. 
where  1-f  m  may  designate  any  number  whatever. 

428.  To  render  the  Logarithmic  Series  converging.  —  The  for 
mula  of  Art.  425, 

fm         /YD 

m--+—  ,  etc.),  (1.) 


can  not  be  employed  for  the  computation  of  logarithms  when 
m  is  greater  than  unity,  because  the  series  does  not  converge. 
This  series  may,  however,  be  transformed  into  a  converging 
series  in  the  following  manner  : 


304  ALGEBRA. 

Substitute  —  m  for  m,  and  we  shall  have 

Subtracting  Eq.  (2)  from  Eq.  (1),  observing  that  log.  (l  +  m] 
— log.  (1— m)  =  log.  -    — ,  we  shall  have 

J.  Tfi 

1  +  ra     ^r/      .  m3  .  m5  . 


Now,  since  this  is  true  for  every  value  of  m,  put 

1 
m  —  —   —  r,  whence 

' 


—  —  , 
1—  m       p 

and  the  preceding  series,  by  substitution,  becomes 
log.  2^=Iog.(p+1)_log./)=2M^+__L_+^:_+,  et,); 

429.  This  series  converges  rapidly,  and  may  be  employed 
for  the  computation  of  logarithms  in  the  Naperian  or  the  com 
mon  systems.  It  is  only  necessary  to  compute  the  logarithms 
of  prime  numbers  directly,  since  the  logarithm  of  any  other 
number  may  be  obtained  by  adding  the  logarithms  of  its  sev 
eral  factors.  Making  £>  =  !,  2,  4,  6,  etc.,  successively,  we  obtain 
the  following 

Naperian  or  Hyperbolic  Logarithms. 

=°-693147 


=1.098612 


log.  4  =  2  log.  2  =1.886294 


log.  6=log.  3-f  log.  2  =1.791769 


log.  8  =  3  log.  2  =2.079442 

log.  9  =  2  log.  3  =2.197225 

log.  10= log.  5 -flog.  2  =2.302585 
etc.,             etc.,  etc. 


LOGARITHMS.  305 

430.  To  construct  a  Table  of  Common  Logarithms.  —  In  order 
to  compute  logarithms  of  the  common  system,  we  must  first 
determine  the  value  of  the  modulus.  In  Art  427,  we  found 

M=     1°g-(1+m) 
Nap.  log.  (1  +  ra)' 

If  ~L-{-m=a,  the  base  of  the  system,  then  log.  a=l,  and  we 
have 

M=Af—  n  -  5 
Nap.  log.  a 

that  is,  the  modulus  of  any  system  is  the  reciprocal  of  the  Naperian 
logarithm  of  the  base  of  the  system. 

The  base  of  the  common  system  is  10,  whose  Naperian  log 
arithm  is  2.302585.  Hence 


_ 

which  is  the  modulus  of  the  common  system. 

We  can  now  compute  the  common  logarithms  by  multiply 
ing  the  corresponding  Naperian  logarithms  by  .434294,  Art. 
427.  In  this  manner  was  the  table  on  pages  290-1  computed. 

431.  Results.  —  The  base  of  Briggs's  system  is       10. 

"  Napier's         "  2.71828. 

The  modulus  of  Briggs's  system  is  0.43429. 

"  Napier's         "        1. 

Since,  in  Briggs's  system,  all  numbers  are  to  be  regarded  as 
powers  of  10,  we  have 


10°-602=4,  etc. 

In  Napier's  system,  all  numbers  are  to  be  regarded  as  powers 
.71828.     Thus, 

2.7180.603^2, 
2.7  IS1-098^  3, 
2.7181-886=4,  etc. 


306  ALGEBKA. 


CHAPTER  XXL 

GENERAL   THEOliY   OF   EQUATIONS. 

432.  A  cubic  equation  with  one  unknown  quantity  is  an 
equation  in  which  the  highest  power  of  this  quantity  is  of  the 
third  degree,  as,  for  example,  x3—  -6x2+8x—  15  =  0.  All  equa 
tions  of  the  third  degree  with  one  unknown  quantity  may  be 
reduced  to  the  form 


A  biquadratic  equation  with  one  unknown  quantity  is  an 
equation  in  which  the  highest  power  of  this  quantity  is  of  the 
fourth  degree,  as,  for  example,  x4  —  6x34-7cc2  +  5x—  4  =  0.  Every 
equation  of  the  fourth  degree  with  one  unknown  quantity  may 
be  reduced  to  the  form 


The  general  form  of  an  equation  of  the  fifth  degree  with  one 
unknown  quantity  is 

x5  +  ax*  +  bx3  +  ex2  +  dx  -f  e  =  0  ; 

and  the  general  form  of  an  equation  of  the  nth  degree  with  one 
unknown  quantity  is 

xn  +  Axn~l  +  'Bxn-*+Cxn-5-{-  ....     +Tx+Y=±;0.       (1.) 

This  equation  will  be  frequently  referred  to  hereafter  by  the 
name  of  the  general  equation  of  the  nth  degree,  or  simply  as 
Equation  (1). 

An  equation  not  given  in  this  form  may  be  reduced  to  it  by 
transposing  all  the  terms  to  the  first  member,  arranging  them 
according  to  the  descending  powers  of  the  unknown  quantity, 
and  dividing  by  the  coefficient  of  the  first  term.  In  this  equa 
tion  n  is  a  positive  whole  number,  but  the  coefficients  A,  B,  C, 
etc.,  may  be  either  positive  or  negative,  entire  or  fractional, 
rational  or  irrational,  real  or  imaginary.  The  term  "V  may  be 
regarded  as  the  coefficient  of  a?°,  and  is  called  the  absolute  term 
of  the  equation. 

It  is  obvious  that  if  we  could  solve  this  equation  we  should 


GENERAL  THEORY  OF  EQUATIONS.         307 

have  the  solution  of  every  equation  that  could  be  proposed. 
Unfortunately,  no  general  solution  has  ever  been  discovered ; 
yet  many  important  properties  are  known  which  enable  us  to 
solve  any  numerical  equation. 

433.  Any  expression,  either  numerical  or  algebraic,  real  or 
imaginary,  which,  being  substituted  for  x  in  Equation  (1),  will 
satisfy  it,  that  is,  make  the  two  members  equal,  is  called  a  root 
of  the  equation. 

It  is  assumed  that  Eq.  (1)  has  at  least  one  root ;  for,  since  the 
first  member  is  equal  to  zero,  it  will  be  so  for  some  value  of 
,T,  either  real  or  imaginary,  and  this  value  of  x  is  by  definition 
a  root. 

434.  If  a  is  a  root  of  the  general  equation  of  the  nth  degree,  its 
first  member  can  be  exactly  divided  by  x — a. 

For  we  may  divide  the  first  member  by  x— a,  according  to 
the  usual  rule  for  division,  and  continue  the  operation  until  a 
remainder  is  found  which  does  not  contain  x.  Let  Q  denote 
the  quotient,  and  R  the  remainder,  if  there  be  one.  Then  we 
shall  have 

^+A:rn-1+Bxw-2_|_....     +To3+V  =  Q(a?-a)  +  R      (2.) 

Now,  if  a  is  a  root  of  the  proposed  equation,  it  will  reduce 
the  first  member  of  (2)  to  0;  it  will  also  reduce  Q(#— a)  to  0; 
hence  R  is  also  equal  to  0.  But,  by  hypothesis,  R  does  no1 
contain  x;  it  is  therefore  equal  to  0,  whatever  value  be  attrib 
uted  to  x,  and,  consequently,  the  first  member  is  exactly  di 
visible  by  x— a. 

435.  If  the  first  member  of  the  general  equation  of  the  nth  de 
gree  is  exactly  divisible  by  x— a,  then  a  is  a  root  of  the  equation. 

For  suppose  the  division  performed,  and  let  Q  denote  the 
quotient;  then  we  shall  have 

xn+Axn~l  +  'Bxr*-2+  ....     +Tx+ Y=Q(x— a). 

If,  in  this  equation,  we  make  cc  =  a,  the  second  member  re« 
duces  to  0;  consequently  the  first  member  reduces  to  0;  and, 
therefore,  a  is  a  root  of  the  equation. 


308  ALGEBRA. 

EXAMPLES. 

1.  Prove  that  1  is  a  root  of  the  equation 


The  first  member  is  divisible  by  as—  •  1,  and  gives  x2~-5x-\-6  =  Q. 

2.  Prove  that  2  is  a  root  of  the  equation 

x3-x-Q  =  0. 
The  first  member  is  divisible  by  x—  2,  and  gives  x2  +  2x+  3  =  0, 

3.  Prove  that  2  is  a  root  of  the  equation 


4.  Prove  that  4  is  a  root  of  the  equation 


5.  Prove  that  —  1  is  a  root  of  the  equation 

^_38^3  +  210x2+  538x+  289=0. 

6.  Prove  that  —5  is  a  root  of  the  equation 

x5  +  6x4_10a?3-112x2-207x-110:=0. 

7.  Prove  that  3  is  a  root  of  the  equation 


436.  Every  equation  of  the  nth  degree  containing  but  one  un 
known  quantity  has  n  roots  and  no  more. 

Since  the  equation  has  at  least  one  root,  denote  that  root  by 
a;  then  will  the  first  member  be  divisible  by  x—  a,  and  the  quo 
tient  will  be  of  the  form 

05"-!  +  A/a5n-2+B'o5w-3+  ....     -]_T'ff+  V, 
and  the  given  equation  may  be  written  under  the  form 

(x—a)(xn-*  +  A'xn-*  +  ....      +  T'o;  +  Y')  =  0.  (3.) 

Now  equation  (3)  may  be  satisfied  by  supposing  either  of  its 
factors  equal  to  zero.  If  the  second  factor  equals  zero,  we 
shall  have 

xn~l+  AV>-2-j-B'xn-3+  ....     +  T'o;+  V'  =  0.         (4.) 

Kow  equation  (4)  has  at  least  one  root  ;  denote  that  root  by 
I;  then  will  the  first  member  be  divisible  by  x—  6,  and  equation 
(4)  can  be  written  under  the  form 


which  reduces  Eq.  (3)  to  the  form  of 


GENERAL  THEORY  OF  EQUATIONS.          309 

By  continuing  this  process,  it  may  be  shown  that  the  first 
member  will  ultimately  be  resolved  into  n  binomial  factors  of 
the  form  x  —  a,  x—b,  x—c,  etc.  Hence  equation  (1)  may  b^ 
written  under  the  form 

(x-a)(x-b)(x-c)(x-d)..  .  .     (x-k)(x-l)  =  0.        (5.) 

This  equation  may  be  satisfied  by  any  one  of  the  n  values, 
x=a,  x=bj  x=Cj  etc.,  and,  consequently,  these  values  are  the 
roots  of  the  equation. 

The  equation  has  no  more  than  n  roots,  because  if  we  ascribe 
to  x  a  value  which  is  not  one  of  the  n  values  a,  6,  c,  etc.,  this 
value  will  not  cause  any  one  of  the  factors  of  Eq.  (5)  to  be  zero, 
and  the  product  of  several  factors  can  not  be  zero  when  neither 
of  the  factors  is  zero. 

If  both  members  of  Eq.  (5)  be  divided  by  either  of  the  fac 
tors  x  —  a,  x—b,  etc.,  it  will  be  reduced  to  an  equation  of  the 
next  inferior  degree;  and  if  we  can  depress  any  equation  to  a 
quadratic,  its  roots  can  be  determined  by  methods  already  ex 
plained. 

Ex.  1.  One  root  of  the  equation 


is  1  ;  what  are  the  other  roots  ? 
Ex.  2.  Two  roots  of  the  equation 


are  1  and  3  ;  what  are  the  other  roots? 
Ex.  3.  Two  roots  of  the  equation 


are  3  and  5  ;  what  are  the  other  roots?  Ans.  2±  V3. 

Ex.  4.  Two  roots  of  the  equation 


are  2  and  3  ;  what  are  the  other  roots?  —  3±  A/5 

Ans.  -  -  —  •  —  . 

4 

Ex.  5.  Two  roots  of  the  equation 

x4-6x3  +  24£-16  =  0 
are  2  and  —2  ;  what  are  the  other  roots?  Ans.  3±  VS. 

437.  The  n  r6ots  of  an  equation  of  the  nth  degree  are  not 
necessarily  all  different  from  each  other.     Any  number,  and,  in- 


310  ALGEBRA.     , 

deed,  all  of  them,  may  be  equal.  When  we  s&y  that  an  equation 
of  the  nth  degree  has  n  roots,  we  simply  mean  that  its  first 
member  can  be  resolved  into  n  binomial  factors,  equal  or  un 
equal,  and  each  factor  contains  one  root. 

Thus  the  equation  x3— 6x2+12x— 8  =  0  can  be  resolved  into 
the  factors  (x-2)(x-2)(x-2)  =  0,  or  (x-2)3  =  0;  whence  it 
appears  that  the  three  roots  of  this  equation  are  2,  2,  2.  But, 
in  general,  the  several  roots  of  an  equation  differ  from  each 
other  numerically. 

The  equation  x3  =  8  has  apparently  but  one  root,  viz.,  2,  but 
by  the  method  of  the  preceding  article  we  can  discover  two 
other  roots.  Dividing  x3 — 8  by  x — 2,  we  obtain  x2-±-2x-\-4:  =  0. 
Solving  this  equation,  we  find  x=  — 1±  V— 3.  Thus,  the  three 
roots  of  the  equation  x3  — 8  are 

2;   _l  +  -v/^3;   -1-V^3. 

The  student  should  verify  the  last  two  values  by  actual  mul 
tiplication. 

Ex.  1.  Find  the  four  roots  of  the  equation  x4— 81  =  0. 
Ex.  2.  Find  the  six  roots  of  the  equation  x6— 64=0. 

438.  The  coefficient  of  the  second  term  in  the  equation  of  the  nth 
degree  is  equal  to  the  algebraic  sum  of  the  roots  with  their  signs 
changed. 

The  coefficient  of  the  third  term  is  equal  to  the  algebraic  sum  of 
the  products  of  all  the  roots,  taken  in  sets  of  two. 

The  coefficient  of  the  fourth  term  is  equal  to  the  algebraic  sum  of 
the  products  of  all  the  roots,  taken  in  sets  of  three,  with  their  signs 
changed. 

The  last  term  is  equal  to  the  continued  product  of  all  the  roots 
with  their  signs  changed. 

Let  a,b,c,d,....  I,  represent  the  roots  of  an  equation  of  the 
nih  degree.  This  equation  will  accordingly  contain  the  factors 
x—a,  x—b,  etc.;  that  is,  we  shall  have 

(x-a)(x-b}(x-c)(x-d) ....     (£c-Z)  =  0. 

If  we  perform  the  multiplication  as  in  Art.  351,  we  shall 
have 


GENERAL  THEORY  OF  EQUATIONS. 


311 


xn— a 

-b 

—  c 

-d 

etc. 


+  bd 


etc. 


—  abd 

—  acd 
-bed 

etc. 


±(abcd....    l)= 


which  results  are  seen  to  conform  to  the  laws  above  stated.  By 
the  method  employed  in  Art.  352  it  may  be  proved  that  if  these 
laws  hold  true  for  the  product  of  n  binomial  factors,  they  will 
also  hold  true  for  the  product  of  ?i  +  l  binomial  factors.  But 
we  have  found  by  actual  multiplication  that  these  laws  are  true 
for  the  product  of  four  factors,  hence  they  are  true  for  the 
product  of  five  factors.  Being  true  for  five,  they  must  be  true 
for  six,  and  so  on  for  any  number  of  factors. 

It  will  be  perceived  that  these  properties  include  those  of 
quadratic  equations  mentioned  on  pages  203-5. 

If  the  roots  are  all  negative,  the  signs  of  all  the  terms  of  the 
equation  will  be  positive,  because  all  the  signs  of  the  factors 
of  which  the  equation  is  composed  are  positive. 

If  the  roots  are  all  positive,  the  signs  of  the  terms  will  be 
alternately  positive  and  negative. 

If  the  sum  of  the  positive  roots  is  numerically  equal  to  the 
sum  of  the  negative  roots,  their  algebraic  sum  will  be  zero; 
consequently  the  coefficient  of  the  second  term  of  the  equation 
will  be  zero,  and  that  term  will  disappear  from  the  equation. 
Conversely,  if  the  second  term  of  the  equation  is  wanting,  the 
sum  of  the  positive  roots  is  numerically  equal  to  the  sum  of 
the  negative  roots. 

Ex.  1.  Form  the  equation  whose  roots  are  1,  2,  and  3. 

For  this  purpose  we  must  multiply  together  the  factors 
Ta;—l?  x — 2,  x— 3,  and  we  obtain  x3  —  6x2  +  llx— 6  =  0. 

This  example  conforms  to  the  rules  above  given  for  the  co 
efficients.  Thus  the  coefficient  of  the  second  term  is  equal  to 
the  sum  of  all  the  roots,  1  +  2  +  3,  with  their  signs  changed. 

The  coefficient  of  the  third  term  is  the  sum  of  the  products 
of  the  roots  taken  two  and  two;  thus, 

1x2  +  1x3  +  2x3  =  11. 


312  ALGEBRA. 

The  last  term  is  the  product  of  all  the  roots,  1x2x3,  with 
their  signs  changed. 

Ex.  2.  Form  the  equation  whose  roots  are  2,  3,  5,  and  —  6. 

Ans.  cc4-4x3-29;r2  +  lt>6x-180  =  0. 

Show  how  these  coefficients  conform  to  the  laws  above  given. 
Ex.  3.  Form  the  equation  whose  roots  are 

1,  1,  1,  -1,  and  -2. 

Ex.  4.  Form  the  equation  whose  roots  are 
1,  3,  5,  -2,  -4,  and  -6. 
Ans.  £6  +  3£5-41x4-87x3+400 
Ex.  5.  Form  the  equation  whose  roots  are 

l±v^2  and  2±V^3. 
Ex.  6.  Form  the  equation  whose  roots  are 
^l  and  2±V3. 


439.  Since  the  last  term  is  the  continued  product  of  all  the 
roots  of  an  equation,  it  must  be  exactly  divisible  by  each  of  them. 

For  example,  take  the  equation  x3  —  x—  6  =  0.  Its  roots  must 
all  be  divisors  of  the  last  term,  6  ;  hence,  if  the  equation  has  a 
rational  root,  it  must  be  one  of  the  numbers  1,  2,  3,  or  6,  either 
positive  or  negative  ;  and,  by  trial,  we  can  easily  ascertain 
whether  either  of  these  numbers  will  satisfy  the  equation.  We 
thus  find  that  +2  is  one  of  the  roots,  and,  by  the  method  of 
Art.  436,  we  find  the  remaining  roots  to  be  —  1±  V  —  2. 

If  the  last  term  of  an  equation  vanishes,  as  in  the  example 
^4  +  2x3  +  3x2  +  6x  =  0,  the  equation  is  divisible  by  x—  0,  and 
consequently  0  is  one  of  its  roots.  If  the  last  two  terms  van 
ish,  then  two  of  its  roots  are  equal  to  zero. 

440.  If  the  coefficients  of  an  equation  are  whole  numbers,  and 
ike  coefficient  of  its  first  term  unity,  the  equation  can  not  have  a 
root  which  is  a  rational  fraction. 

Suppose,  if  possible,  that  T  is  a  root  of  the  general  equation 

of  the  nth.  degree,  where  j-  represents  a  rational  fraction  ex 

pressed  in  its  lowest  terms.     Substitute  this  value  for  x  in  the 
given  equation,  and  we  have 


GENERAL  THEORY  OF  EQUATIONS.          313 


Multiplying  each  term  by  bn~\  and  transposing,  we  obtain 


Now,  by  supposition,  a,  6,  A,  B,  C,  etc.,  are  whole  numbers; 
hence  the  right-hand  member  of  the  equation  is  a  whole  num 
ber. 

But,  by  hypothesis,  |  is  an  irreducible  fraction  ;  that  is,  a 
and  b  contain  no  common  factor.  Consequently,  an  and  b  will 
contain  no  common  factor;  that  is,  ~  is  a  fraction  in  its  low 
est  terms.  Hence  the  supposition  that  the  irreducible  fraction 
7  is  a  root  of  the  equation  leads  to  this  absurdity,  that  an  irre 

ducible  fraction  is  equal  to  a  whole  number. 

This  proposition  only  asserts  that  every  commensurable  root 
must  be  an  integer.  The  roots  can  not  be  of  the  form  of  -|,  |-, 
•f,  etc.  The  equation  may  have  other  roots  which  are  incom 
mensurable  or  imaginary,  as  2±V3,  1  ±  V  —  2. 

. 

441.  Any  equation  having  fractional  coefficients  can  be  trans* 
formed  into  another  which  has  alt  its  coefficients  integers,  and  the 
coefficient  of  its  first  term  unity. 

Reduce  the  equation  to  the  form 

Axn  +  'Bzn-l  +  Cxn-*+  ____  -fTcc-fV=0, 
in  which  A>  B,  C,  etc.,  are  all  integers,  either  positive  or  neg 
ative. 

Substitute  fur  x  the  value   x=-r-, 

A. 

and  the  equation  becomes 


which,  multiplied  by  A71"1,  becomes 


2/n-f  By'1-1  +  AC?/'1-2  +  ....   -f  A'l-2T?/4-  A'^VrrrO. 

O 


814  ALGEBKA. 

in  which  the  coefficients  are  all  integers,  and  that  of  the  first 
term  unity. 

The  substitution  of  ~  for  x  is  not  always  the  one  which  leads 

to  the  most  simple  result  ;  but  when  A  contains  two  or  more 
equal  factors,  each  factor  need  scarcely  ever  be  repeated  more 
than  once. 

^x2     5x     2 
Ex.  1.  Transform  the  equation  x3  —  —-\  —  -  —  -=  0  into  an- 

A      4     y 

other  whose  coefficients  are  integers,  and  that  of  the  first  term 
unity. 

Clearing  of  fractions,  we  have 


?y 
Substituting  ^  for  x,  the  transformed  equation  is 

y3     9?/2     45y     R   _0 
6~-~6""~6~-          =°J 

or  ?/3-9?/2+45?/-48  =  0. 

Transform  the  following  equations  into  others  whose  coeffi 
cients  are  integers,  and  that  of  the  first  term  unity. 

Ex.2.   x3  +  2x2  +  |-^0.         Ans.  ?/3-hl2?/2-f9?/-24  =  0. 
Ex.3.   x3+°?—+2  =  0. 


Ex.4. 
Ex.5. 
Ex.6. 


. 

o 


442.  If  in  any  complete  equation  involving  but  one  unknown 
quantity  the  signs  of  the  alternate  terms  be  changed,  the  signs  of  all 
the  roots  will  be  changed. 

Take  the  general  equation  of  the  nth  degree, 

xn  +  Ax"-1  +  Bxn-2  +  (V*-3+  ....     =o.          (1.) 
in  which  the  signs  may  follow  each  other  in  any  order  whatever. 


GENERAL  THEORY  OF  EQUATIONS.          315 

If  we  change  the  signs  of  the  alternate  terms,  we  shall  have 

xn  —  Axn~l  +  ~Bxn~*  —  Gxn~s  4-  ____     =0.         (2  .) 
or,  changing  the  sign  of  every  term  of  the  last  equation, 

—  xn  4-  Axn~l  —  Bxn~2  +  Cxn~*  —  ____      =0.         (3.  ) 

Now,  substituting  +a  for  x  in  equation  (1)  will  give  the 
same  result  as  substituting  —a  in  equation  (2),  if  n  be  an  even 
number;  or  substituting  —a  in  equation  (3),  if  n  be  an  odd 
number.  If,  then,  a  is  a  root  of  equation  (1),  —a  will  be  a  root 
of  equation  (2),  and,  of  course,  a  root  of  equation  (3),  which  is 
identical  with  it. 

Hence  we  see  that  the  positive  roots  may  be  changed  into 
negative  roots,  and  the  reverse,  by  simply  changing  the  signs 
of  the  alternate  terms  ;  so  that  the  finding  the  real  roots  of  any 
equation  is  reduced  to  finding  positive  roots  only. 

This  rule  assumes  that  the  proposed  equation  is  complete; 
that  is,  that  it  has  all  the  terms  which  can  occur  in  an  equation 
of  its  degree.  If  the  equation  be  incomplete,  we  must  intro 
duce  any  missing  term  with  zero  for  its  coefficient. 

Ex.  1.  The  roots  of  the  equation  x3  —  2x2—5x-\-Q—0  are  1, 
3,  and  —2  ;  what  are  the  roots  of  the  equation 


Ex.  2.  The  roots  of  the  equation  x3  —  6x24  lice—  6  =  0  are  1, 
2}  and  3  ;  what  are  the  roots  of  the  equation 


Ex.  3.  The  roots  of  the  equation  x4  —  6x3  +  5x2  -f2#—  10  =  0 
are  —1,  +5,  1  +  V—  1,  and  1—  V  —  1  ;  what  are  the  roots  of 
the  equation  x4  -f  6x3  -f  5x2  —  2x  —  10  =  0  ? 

443.  If  an  equation  whose  coefficients  are  all  real  contains  im 
aginary  roots,  the  number  of  these  roots  must  be  even. 

If  an  equation  whose  coefficients  are  all  real  has  a  root  of 
the  form  a  +  bV  —  1,  then  will  a  —  bV  —  1  be  also  a  root  of  the 
equation.  For,  let  a  +  bV  —  1  be  substituted  for  x  in  the  equa 
tion,  the  result  will  consist  of  a  series  of  terms,  of  which  those 
involving  only  the  powers  of  a  and  the  even  powers  of  bV  —  I 
will  be  real,  and  those  which  involve  the  odd  powers  o 
will  be  imaginary. 


316  ALGEBKA. 


If  we  denote  the  sum  of  the  real  terras  by  P,  and  the  sum 
of  the  imaginary  terms  by  Q\/  —  1,  the  equation  becomes 
P+QV^7=0. 

But,  according  to  Art.  243,  this  equation  can  only  be  true 
when  we  have  separately  P  —  0  and  Q  =  0. 

If  we  substitute  a  —  bV  —  l  for  x  in  the  proposed  equation, 
the  result  will  differ  from  the  preceding  only  in  the  signs  of 
the  odd  powers  of  b  V—  1,  so  that  the  result  will  be  P—  QV^l. 


But  we  have  found  that  P  =  0  and  Q=0  ;  hence  P-Q\/37[  ^0. 
Therefore  a—  bV  —  1,  when  substituted  for  x,  satisfies  the  equa 
tion,  and,  consequently,  it  is  a  root  of  the  equation. 

It  may  be  proved  in  a  similar  manner  that  if  an  equation 
whose  coefficients  are  all  rational,  has  a  root  of  the  form  a-\-  Vb, 
then  will  a—  Vb  be  also  a  root  of  the  equation. 

Ex.1.  One  root  of  the  equation  x3—  2x-j-4  =  0  is  1  +  V  —  1  ; 
what  are  the  other  roots? 

Ex.  2.  One  root  of  the  equation  x3  —  x2  —  7x  -f  15  =  0  is 
2+  V—  1  ;  what  are  the  other  roots? 

Ex.  3.  One   root    of  the   equation    x3  —  xz  +  3x  +  5  =  0    is 

1  +  2V  —  1  ,  what  are  the  other  roots? 

Ex.  4.  One  root  of  the   equation   x4  —  4x3  +  4x  —  1  —  0  is 

2  +  A/3  ;  what  are  the  other  roots  ? 

Ex.  5.  Two  roots  of  the  equation 


are  _l_|_-y/_i  and  1—  V—  3;  what  are  the  other  six  roots? 

444.  Any  equation  involving  but  one  unknown  quantity  may  be 
transformed  into  another  whose  roots  differ  from  those  of  the  pro 
posed  equation  by  any  given  quantity. 

Let  it  be  required  to  transform  the  general  equation  of  the 
nth  degree  into  another  whose  roots  shall  be  less  than  those  of 
the  proposed  equation  by  a  constant  difference  h. 

Assume  y—x—li,  whence  x  =  y+h. 

Substituting  y-\-h  for  x  in  the  proposed  equation,  we  have 


GENERAL  THEORY  OF  EQUATIONS,          317 

Developing  the  different  powers  of  y  +  h  by  the  binomial 
formula,  and  arranging  according  to  the  powers  of  y,  we  have 
if+nhyn~l 


+  (n-I)Ah 
+  B 


+  (w-2)B// 


+  C 

which  equation  satisfies  the  proposed  condition,  since  y  is  less 
than  x  by  h.  If  we  assume  y  =  x-{-h,  or  x=y—h,  we  shall  ob 
tain  in  the  same  manner  an  equation  whose  roots  are  greater 
than  those  of  the  given  equation  by  h. 

Ex.  1.  Find  the  equation  whose  roots  are  greater  by  1  than 
those  of  the  equation  x3-\-Sxz— 4x+l  =  0. 
We  must  here  substitute  y—l  in  place  of  x. 

Ans.  ?/3_7?/-(-7  — 0. 

Ex.  2.  Find  the  equation  whose  roots  are  less  by  1  than  those 
of  the  equation  x3— 2o?2+3as— 4=0. 

Ans.  y3  +  y2  +  2y—2  =  Q. 

Ex.  3.  Find  the  equation  whose  roots  are  greater  by  3  than 
those  of  the  equation 'x4  +  9 

Ans. 

Ex.  4.  Find  the  equation  whose  roots  are  less  by  2  than  those 
of  the  equation  5x4  — 12x3  +  3x2  +  4^-5  =  0. 

Ans.  5?/4  +  28?/3  +  51?/2+32?/-l  =  0. 

Ex.  5.  Find  the  equation  whose  roots  are  greater  by  2  than 
those  of  the  equation  x5  +  10x4H-4 

Ans. 

445.  Any  complete  equation  may  be  transformed  into  another 
wliose  second  term  is  wanting. 

Since  h  in  the  preceding  article  may  be  assumed  of  any 
value,  we  may  put  nh  +  A  =  0,  which  will  cause  the  second  term 

of  the  general  development  to  disappear.     Hence  li— -  and 

A  w' 

x  =  y — -.     Hence,  to  transform  an  equation  into  another  which 

?& 

wants  the  second  term,  substitute  for  the  unknown  quantity  a  new 
unknown  quantity  minus  the  coefficient  of  the  second  term  divided 
ly  the  highest  exponent  of  the  unknown  quantity. 


318  ALGEBRA. 

Ex.l.  Transform  the  equation  x3—  6x2  +  8x  —  2  =  0  into  an 
other  whose  second  term  is  wanting. 

Put  #=2/4-2.  Ans.  y3—4y—2  =  0. 

Ex.2.  Transform  the  equation  x4—l6x3—6x-}-l6~0  into 
another  whose  second  term  is  wanting. 

Put  sc=2/+4.  Ans.  y4- 

Ex.  3.  Transform  the  equation 


into  another  whose  second  term  is  wanting. 

Ans.  ?/5-78?/3+412?/2-757?/-h401=0. 

Ex.4.  Transform  the  equation  x4—  8x3-f5  =  0  into  another 
whose  second  term  is  wanting. 

According  to  Art.  438,  when  the  second  term  of  an  equation 
is  wanting,  the  sum  of  the  positive  roots  is  numerically  equal 
to  the  sum  of  the  negative  roots. 

446.  If  two  numbers,  substituted  for  tht  unknown  quantity  in  an 
equation,  give  results  with  contrary  signs,  there  must  be  at  least  one 
real  root  included  between  those  numbers. 

Let  us  denote  the  real  roots  of  the  general  equation  of  the 
nth  degree  by  a,  I,  c,  etc.,  and  suppose  them  arranged  in  the 
order  of  their  magnitude,  a  being  algebraically  the  smallest, 
that  is,  nearest  to  —  a  ;  b  the  next  smallest,  and  so  on.  The 
equation  may  be  written  under  the  following  form, 
(x—a)(x—  b)(x—  c)(x-d)  .....  =0. 

Now  let  us  suppose  x  to  increase  from  —  a  toward  4-  a  , 
assuming,  in  succession,  every  possible  value.  As  long  as  x  is 
less  than  a,  every  factor  of  the  above  expression  will  be  nega 
tive,  and  the  entire  product  will  be  positive  or  negative  accord 
ing  as  the  number  of  factors  is  even  or  odd.  When  x  becomes 
equal  to  a,  the  whole  product  becomes  equal  to  0.  But  if  x  be 
greater  than  a  and  less  than  5,  the  factor  x—  a  will  be  positive, 
while  all  the  other  factors  will  be  negative.  Hence,  when  x 
changes  from  a  value  less  than  a  to  a  value  greater  than  a  and 
less  than  6,  the  sign  of  the  whole  product  changes  from  -h  to  — 
or  from  —  to  -f-.  When  x  becomes  equal  to  b,  the  product 
again  becomes  zero  ;  and  as  x  increases  from  b  to  c,  the  factor 


GENERAL  THEORY  OF  EQUATIONS.          319 

#_&  becomes  positive,  and  the  sign  of  the  product  changes 
again  from  —  to  -f  or  from  +  to  —  ;  and,  in  general,  the  prod 
uct  changes  its  sign  as  often  as  the  value  of  x  passes  over  a  real 
root  of  the  equation. 

Hence,  if  two  numbers  substituted  for  x  in  an  equation  give 
results  with  contrary  signs,  there  must  be  some  intermediate 
number  which  reduces  the  first  member  to  0,  and  this  number 
is  a  root  of  the  equation. 

If  the  two  numbers  which  give  results  with  contrary  signs 
differ  from  each  other  only  by  unity,  it  is  plain  that  we  have 
found  the  integral  part  of  a  root. 

If  two  numbers,  substituted  for  x  in  an  equation,  give  results 
with  like  signs,  then  between  these  numbers  there  will  either 
be  no  root,  or  some  even  number  of  roots.  The  last  case  may 
include  imaginary  roots. 

Forifa  +  6'/—  1  be  a  root  of  the  equation,  then  will  a  —  IV  —  1 
be  also  a  root.  Now 


a  result  which  is  always  positive;  that  is,  the  quadratic  factor 
corresponding  to  a  pair  of  imaginary  roots  of  an  equation  whose 
coefficients  are  real,  is  always  positive. 

Ex.  1.  Find  the  first  figure  of  one  of  the  roots  of  the  equa 
tion  xt+xz+x—  100  =  0. 

When  x=41  the  first  member  of  the  equation  reduces  to 
—  16;  and  when  05  =5,  it  reduces  to  -f  55.  Hence  there  must 
be  a  root  between  4  and  5  ;  that  is,  4  is  the  first  figure  of  one 
of  the  roots. 

Ex.  2.  Find  the  first  figure  of  one  of  the  roots  of  the  equa 
tion  cc3-6x2  +  9x-10  =  0. 

Ex.  3.  Find  the  first  figure  of  each  of  the  roots  of  the  equa 
tion  x3—  ±x2— 


447.  In  a  series  of  terms,  two  successive  signs  constitute  a 
permanence  when  the  signs  are  alike,  and  &  variation  when  they 
are  unlike.  Thus,  in  the  equation  x3  —  2x2  —  5#+6  =  0,  the  signs 
of  the  first  two  terms  constitute  a  variation,  the  signs  of  the 
second  and  third  constitute  a  permanence,  and  those  of  the 
third  and  fourth  also  a  variation. 


320  ALGEBRA. 

Descartes' s  Rule  of  Signs. 

448.  Every  equation  must  have  as  many  variations  of  sign  as 
it  has  positive  roots,  and  as  many  permanences  of  sign  as  it  has 
negative  ?*oots. 

According  to  Art.  436,  the  first  member  of  tbe  general  equa 
tion  of  the  Tith  degree  may  be  regarded  as  the  prcrduct  of  n  bi 
nomial  factors  of  the  form  x—a,x  —  b,  etc.  The  above  theorem 
will  then  be  demonstrated  if  we  prove  that  the  multiplication 
of  a  polynomial  by  a  new  factor,  x — a,  corresponding  to  a  pos 
itive  root,  will  introduce  at  least  one  variation,  and  that  the  mul 
tiplication  by  a  factor,  x+a,  will  introduce  at  least  one  perma* 
nence. 

Suppose,  for  example,  that  the  signs  of  the  terms  in  the 

original  polynomial  are  -M --\ 1 1- ,  and  we  have 

to  multiply  the  polynomial  by  a  binomial  in  which  the  signs 
of  the  terms  are  -| — .  If  we  write  down  simply  the  signs 
which  occur  in  the  process  and  in  the  result,  we  have 


±  - 


"We  perceive  that  the  signs  in  the  upper  line  of  the  partial 
products  must  all  be  the  same  as  in  the  given  polynomial  ;  but 
those  in  the  lower  line  are  all  contrary  to  those  of  the  given 
polynomial,  and  advanced  one  term  toward  the  right.  When 
the  corresponding  terms  of  the  two  partial  products  have  dif 
ferent  signs,  the  sign  of  that  term  in  the  result  will  depend 
upon  the  relative  magnitude  of  the  two  terms,  and  may  be 
either  -f  or  —  .  Such  terms  have  been  indicated  by  the  double 
sign  ±  ;  and  it  will  be  observed  that  the  permanences  in  the 
given  polynomial  are  changed  into  signs  of  ambiguity.  Hence, 
take  the  ambiguous  sign  as  you  will,  the  permanences  in  the 
final  product  are  not  increased  by  the  introduction  of  the  posi 
tive  root  -fa,  but  the  number  of  signs  is  increased  by  one,  ancl 


GENERAL  THEORY  OF  EQUATIONS. 


321 


therefore  the  number  of  variations  must  be  increased  by  one. 
Hence  each  factor  corresponding  to  a  positive  root  must  intro 
duce  at  least  one  new  variation,  so  that  there  must  be  as  many 
variations  as  there  are  positive  roots. 

In  the  same  manner  we  may  prove  that  the  multiplication 
by  a  factor,  x  +  a,  corresponding  to  a  negative  root,  must  intn> 
duce  at  least  one  new  permanence  ;  so  that  there  must  be  as 
many  permanences  as  there  are  negative  roots. 

If  all  the  roots  of  an  equation  are  real,  the  number  of  posi 
tive  roots  is  equal  to  the  number  of  variations,  and  the  number 
of  negative  roots  is  equal  to  the  number  of  permanences.  If 
the  equation  is  incomplete,  we  must  supply  the  place  of  any 
deficient  term  with  ±0  before  applying  the  preceding  rule. 

Ex.1.  The  equation  x5-3x4-5x3  +  15^2+4x-12  =  0  has 
five  real  roots;  how  many  of  them  are  positive? 

Ex.2.  The  equation  cc4—  3x3—  15x2  +49^-12  =  0  has  four 
real  roots  ;  how  many  of  them  are  negative  ? 

Ex.  3.  The  equation 


has  six  real  roots;  how  many  of  them  are  positive? 

Derived  Polynomials. 

449.  If  we  take  the  general  equation  of  the  nth  degree,  and 
substitute  y  +  h  in  place  of  x,  it  becomes 


Developing  the  powers  of  the  binomial  y+h,  and  arranging 
in  the  order  of  the  powers  of  h,  we  have 

A2 


y 


-i 


1.2 


The  part  of  this  development  which  is  independent  of  h  is 
of  the  same  form  as  the  original  polynomial,  and  we  will  des 

O  2 


322  ALGEBRA. 

ignate  it  by  X.     We  will  denote  the  coefficient  of  h  by  Xu  the 

iiz 
coefficient  of  -^-^  by  X2,  etc-     The  preceding  development  may 

then  be  written 


450.  The  polynomials  X1?  X2,  etc.,  are  called  derived  poly 
nomials,  or  simply  derivatives.  XT  is  called  the  first  derivative 
of  X,  X2  the  second  derivative,  and  so  on.  X  is  called  the  prim 
itive  polynomial.  Each  derived  polynomial  is  deduced  from  the 
preceding  by  multiplying  each  term  by  the  exponent  of  the  leading 
letter  in  that  term,  and  then  diminishing  the  exponent  of  the  lead 
ing  letter  by  unity. 

Ex.  1.  What  are  the  successive  derivatives  of 


(1st.  3x2- 
Ans.  \  2d.  6a- 

(  3d.  6. 
Ex.  2.  What  are  the  successive  derivatives  of 


Ex.  3.  What  are  the  successive  derivatives  of 

x5-f3x4  +  2x3-3x2-2x-2? 
Ex.  4.  What  is  the  first  derivative  of 


+  ____     +Tx+V? 

Equal  Roots. 

451.  We  have  seen,  Art.  436,  that  if  a,  I,  c,  etc.,  are  the  roots 
of  the  general  equation  of  the  nth  degree,  the  equation  may  be 
written 

X  =  (x-a)(x-l)(x-c).  .  .  .      (x-k)(x-l)  =  0. 

When  the  equation  has  two  roots  equal  to  a,  there  will  be 
two  factors  equal  to  x—a;  that  is,  the  first  member  will  be  di 
visible  by  (x—a)2;  when  there  are  three  roots  equal  to  a,  the 
first  member  will  be  divisible  by  (x—a)3;  and  if  there  are  n 
roots  equal  to  a,  the  first  member  will  contain  the  factor  (x—  o)n- 
The  first  derivative  will  contain  the  factor  n(x  —  a)71"1;  that  is, 


GENERAL  THEORY  OF  EQUATIONS.          323 

x—  a  occurs  (n—  1)  times  as  a  factor  in  the  first  derivative. 
The  greatest  common  divisor  of  the  primitive  polynomial,  and 
its  first  derivative,  must  therefore  contain  the  factor  x—a,  re 
peated  once  less  than  in  the  primitive  polynomial. 

Hence,  to  determine  whether  an  equation  has  equal  roots,  we 
have  the  following 

RULE. 

Find  the  greatest  common  divisor  between  the  given  polynomial 
and  its  first  derivative.  If  there  is  no  common  divisor  the  equation 
has  no  equal  roots.  If  there  is  a  common  divisor,  place  this  equal 
to  zero,  and  solve  the  resulting  equation. 

Ex.  1.  Find  the  equal  roots  of  the  equation 


The  first  derivative  is  8xz—  16x-f-21. 

The  greatest  common  divisor  between  this  and  the  given 
polynomial  is  x—  3. 

Hence  the  equation  has  two  roots,  each  equal  to  3. 
Ex.  2.  Find  the  equal  roots  of  the  equation 


Ans.  Two  roots  equal  to  5. 
Ex.  3.  Find  the  equal  roots  of  the  equation 


Ans.  Two  roots  equal  to  2. 
Ex.  4.  Find  the  equal  roots  of  the  equation 
x*-6xz—  8x-3  =  0. 

Ans.  Three  roots  equal  to  —1. 
Ex.  5.  Find  the  equal  roots  of  the  equation 


Ex.  6.  Find  the  equal  roots  of  the  equation 

-j-  20x4-16  =  0. 


/Sturm's  Theorem. 

452.  The  object  of  Sturm's  theorem  is  to  determine  the  number 
of  the  real  roots  of  an  equation,  and  likewise  the  situation  of 
these  roots,  or  their  initial  figures  when  the  roots  are  irrational 

According  to  Art.  446,  if  we  suppose  x  to  assume  in  succes 
sion  every  possible  value  from  —  a  to  +  a,  and  determine  the 


824  ALGEBRA. 

number  of  times  that  the  first  member  of  the  equation  changes 
its  sign,  we  shall  have  the  number  of  real  roots,  and,  conse* 
quently,  the  number  of  imaginary  roots  in  the  equation,  since 
the  real  and  imaginary  roots  are  together  equal  in  number  to 
the  degree  of  the  equation.  Sturm's  theorem  enables  us  easily 
to  determine  the  number  of  such  changes  of  sign. 

453.  Sturm's  Functions. — Let  the  first  member  of  the  general 
equation  of  the  nth  degree,  after  having  been  freed  from  its 
equal  roots,  be  denoted  by  X,  and  let  its  first  derivative  be  de 
noted  by  Xr     We  now  apply  to  X  and  Xx  the  process  of  find 
ing  their  greatest  common  divisor,  with  this  modification,  that 
we  change  the  sign  of  each  remainder  before  taking  it  as  a  di 
visor;  that  is,  divide  X  by  X1?  and  denote  the  remainder  with 
its  sign  changed  by  E;  also,  divide  Xa  by  E,  and  denote  the 
remainder  with  its  sign  changed  by  Ep  and  so  on  to  En,  which 
will  be  a  numerical  remainder  independent  of  x,  since,  by  hy 
pothesis,  the  equation  X=0  has  no  equal  roots. 

We  thus  obtain  the  series  of  quantities 

X,  X1?  E,  Er  E2, ....     En, 

each  of  which  is  of  a  lower  degree  with  respect  to  x  than  the 
preceding;  and  the  last  is  altogether  independent  of  x,  that  is, 
does  not  contain  x. 

We  now  substitute  for  x  in  the  above  functions  any  two 
numbers,  p  and  q,  of  which  p  is  less  than  q.  The  substitution 
of  p  will  give  results  either  positive  or  negative.  If  we  only 
take  account  of  the  signs  of  the  results,  we  shall  obtain  a  cer 
tain  number  of  variations  and  a  certain  number  of  perma 
nences. 

The  substitution  of  q  for  x  will  give  a  second  series  of  signs, 
presenting  a  certain  number  of  variations  and  permanences. 
The  following,  then,  is  the  Theorem  of  Sturm. 

454.  Jfj  in  the  series  of  functions  X,  X15  E,  E15  ....     En,  we 
substitute  in  place  of  x  any  two  numbers,  p  and  q,  either  positive 
or  negative,  and  note  the  signs  of  the  results,  the  difference  between 
the  number  of  variations  of  sign  when  x—p  and  when  x  =  q  is  equal 


GENERAL  THEORY  OF  EQUATIONS.          325' 

to  the  number  of  real  roots  of  the  equation  X  =  0  comprised  between 
p  and  q. 

Let  Q,  Qj,  Q2, .  . . .  Qn,  denote  the  quotients  in  the  success 
ive  divisions.  Now,  since  the  dividend  is  equal  to  the  product 
of  the  divisor  and  quotient  plus  the  remainder,  or  minus  the 
remainder  with  its  sign  changed,  we  must  have  the  following 
aquations : 

X=X1Q-R,  (1.) 

X1  =  RQ,  -R>,  (2.) 

R=R1Q2-RZ,  (3.) 


Rn_2  =  Rn-lQn  —  Rn-  (n  —  1). 

From  these  equations  we  deduce  the  following  conclusions: 

455.  If,  in  the  series  of  functions  X,  Xr  R,  etc.,  any  number  be 
substituted  for  x,  two  consecutive  functions  can  not  reduce  to  zero  at 
the  same  lime. 

For,  if  possible,  suppose  X1  =  0  and  R  =  0;  then,  by  Eq.  (2), 
we  shall  have  R^O.  Also,  since  R=0  and  E1  =  0,  by  Eq.  (3) 
we  must  have  R2  =  0;  and  from  the  next  equation  R3  —  0,  and 
so  on  to  the  last  equation,  which  will  give  Rn=0,  which  is  im 
possible,  since  it  was  shown  that  this  final  remainder  is  inde 
pendent  of  a:,  and  must  therefore  remain  unchanged  for  every 
value  of  x. 

456.  When,  by  the  substitution  of  any  number  for  x.  any  one  of 
these  functions  becomes  zero,  the  two  adjacent  functions  must  have 
contrary  signs  for  the  same  value  of  x. 

For,  suppose  Rj  in  Eq.  (3)  becomes  equal  to  zero,  then  this 
equation  will  reduce  to  R=  — R2;  that  is,  R  and  R2  have  con 
trary  signs. 

457.  If  a  is  a  root  of  the  equation  X  =  0,  the  signs  ofK.  and  X, 
will  constitute  a  variation  for  a  value  of  x  which  is  a  little  less  than 
a,  and  a  permanence  for  a  value  of  x  which  is  a  little  greater  than  a. 

Let  h  denote  a  positive  quantity  as  small  as  we  please,  and 
let  us  substitute  a+h  for  x  in  the  equation  X  =  0.  According 


u2<>  ALGEBRA. 

to  Art.  449,  the  development  will  be  of  the  form 

A2 
X  +  XjA-fX.^  —  +  other  terms  involving  higher  powers  of  h< 

Now,  if  a  is  a  root  of  the  proposed  equation,  it  must  reduce 
the  polynomial  X  to  zero,  and  the  development  becomes 

h2 

—  -j-other  terms  involving  higher  powers  of  A, 

2 


or  7z(X1  +  Xa    +  ,  etc.). 

Zi 

Also,  if  we  substitute  a  -\-  h  for  x  in  the  first  derived  polyno 
mial,  the  development  will  be  of  the  form 

Xj  +  X27i-|-  other  terms  involving  higher  powers  of  h. 

Now  a  value  may  be  assigned  to  h  so  small  that  the  first 
term  of  each  of  these  developments  shall  be  greater  than  the 
sum  of  all  the  subsequent  terms.  For  if  h  be  made  indefinitely 
small,  then  will  X2/i  be  indefinitely  small  in  comparison  with 
Xj,  which  is  finite;  and,  since  the  following  terms  contain 
higher  powers  of  h  than  the  first,  each  will  be  indefinitely  small 
in  comparison  with  the  preceding  term  ;  and,  since  the  number 
of  terms  is  finite,  the  first  term  must  be  greater  than  the  sum 
of  the  subsequent  terms.  Hence,  when  li  is  taken  indefinitely 
small,  the  sum  of  the  terms  of  the  two  developments  must  have 
the  same  sign  as  their  first  terms, 

X^  and  Xr 

When  h  is  positive,  these  terms  must  both  have  the  same 
sign  ;  and  when  li  is  negative,  they  must  have  contrary  signs  ; 
that  is,  the  signs  of  the  two  functions  X  and  Xj  constitute  a 
variation  when  x  —  a—li^  and  a  permanence  when  x=a+h. 

458.  Demonstration  of  Sturm's  Tlieorem.  —  Suppose  all  the  real 
roots  of  the  equations 

X  =  0,  X1  =  0,  K=:0,  R^O,  etc., 

to  be  arranged  in  a  series  in  the  order  of  magnitude,  beginning 
with  the  least.  Let^>  be  less  than  the  least  of  these  roots,  and 
let  it  increase  continually  until  it  becomes  equal  to  q,  which  we 
suppose  to  be  greater  than  the  greatest  of  these  roots.  Now, 
so  long  as  p  is  less  than  any  of  the  roots,  no  change  of  sign  will 


GENERAL  THEORY  OF  EQUATIONS.          327 

occur  from  the  substitution  of  p  for  x  in  any  of  these  functions, 
Art.  446.  But  suppose  j>  to  pass  from  a  number  a  little  small 
er  to  a  number  a  little  greater  than  a  root  of  the  equation  X  =  0, 
the  sign  of  X  will  be  changed  from  +  to  —  or  from  —  to  +, 
Art.  446.  The  signs  of  X  and  Xx  constitute  a  variation  before 
the  change,  and  a  permanence  after  the  change,  Art.  457 ;  that 
is,  there  is  a  variation  lost  or  changed  into  a  permanence. 

Again,  while  p  increases  from  a  number  a  little  smaller  to  a 
number  a  little  greater  than  another  root  of  the  equation  X=0, 
a  second  variation  will  be  changed  into  a  permanence,  and  so 
on  for  the  other  roots  of  the  given  equation. 

But  when  p  arrives  at  a  root  of  any  of  the  other  functions 
X1}  R,  E1}  its  substitution  for  x  reduces  that  polynomial  to 
zero,  and  neither  the  preceding  nor  succeeding  functions  can 
vanish  for  the  same  value  of  x,  Art.  455 ;  and  these  two  adja 
cent  functions  have  contrary  signs,  Art.  456.  Hence  the  en 
tire  number  of  variations  of  sign  is  not  affected  by  the  vanish 
ing  of  any  function  intermediate  between  X  and  En,  for  the 
three  adjacent  functions  must  reduce  to  +0—  or  — 0  +  . 

Here  is  one  variation,  and  there  will  also  be  one  variation 
if  we  supply  the  place  of  the  0  with  either  +  or  —  ;  thus, 
+  ±  —  or  —  ±  +. 

Thus  we  have  proved  that  during  all  the  changes  of  p, 
Sturm's  functions  never  lose  a  variation  except  when  p  passes 
through  a  root  of  the  equation  X  =  0,  and  they  never  gain  a 
variation.  Hence  the  number  of  variations  lost  while  x  in 
creases  from  p  to  q  is  equal  to  the  number  of  the  roots  of  the 
equation. X  — 0,  which  lie  between  _p  and  q. 

Now,  since  all  the  real  roots  must  be  comprised  within  the 
limits  —  oc  and  -f  oc ,  if  we  substitute  these  values  for  x  in  the 
series  of  functions  X,  X15  etc.,  the  number  of  variations  lost 
will  indicate  the  whole  number  of  real  roots.  A  third  suppo 
sition  that  x  —  0  will  show  how  many  of  these  roots  are  posi 
tive  and  how  many  negative ;  and  if  we  wish  to  determine 
smaller  limits  of  the  roots,  we  must  try  other  numbers.  It  is 
generally  best  in  the  first  instance  to  make  trial  of  such  num 
bers  as  are  most  convenient  in  computation,  as  1,  2,  10,  etc. 


328  ALGEBKA. 


EXAMPLES. 

1.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x3 — 3x2 — 4x  +  13  =  0 

Here  we  have  X=x3— 3x2-4^+13,  and  X1=:3x2-6a7~4. 
Dividing  X  by  X17  we  find  for  a  remainder  —  14x-f  35.  Ee- 
jecting  the  factor  7,  and  changing  the  sign  of  the  result,  we 
have  E=2x— 5.  Multiplying  X:  by  4,  and  dividing  by  E,  we 
find  for  a  remainder  —1.  Changing  the  sign,  we  have  Ex=  +1. 

Hence  we  have 


If  we  substitute  —  oc  for  x  in  the  polynomial  X,  the  sign  of 
the  result  is  — ;  if  we  substitute  —  oc  for  x  in  the  polynomial 
X1?  the  sign  of  the  result  is  + ;  if  we  substitute  —  oc  for  x  in 
the  expression  2x— 5,  the  sign  of  the  result  is  — ;  and  E1?  being 
independent  of  x,  will  remain  +  for  every  value  of  x,  so  that, 
by  supposing  x= —  oc,  we  obtain  the  series  of  signs 

+         +• 

Proceeding  in  the  same  manner  for  other  assumed  values  of 
a?,  we  shall  obtain  the  following  results : 

Assumed  Values  of  a:.         Resulting  Signs.  Variations. 

—  oc  +  —  +  giving  3  variations. 

-3  —  +  —  +  "3  " 

-  2  +  +  +  "2 

0  +  -  -     +  "       2  " 

1  +  •  +  "      2  " 

2  +  —  +  "      2  " 
2J  -  0   +  "       1 

3  +  +  +  +  "      0 
+  oc                 +  +  +  +                   "0 

We  perceive  that  no  change  of  sign  in  either  function  occurs 
by  the  substitution  for  x  of  any  number  less  than  — 3  ;  but,  in 
passing  from  —3  to  —2,  the  function  X  changes  its  sign  from 
—  to  +,  by  which  one  variation  is  lost.  In  passing  from  2 


GENERAL  THEORY  OF  EQUATIONS.          329 

to  2J,  the  function  X  again  changes  its  sign,  and  a  second  va 
riation  of  sign  is  lost.  Also,  in  passing  from  2-J  to  3,  the  func 
tion  X  again  changes  its  sign,  and  a  third  variation  is  lost;  and 
there  are  no  further  changes  of  sign  arising  from  the  substitu 
tion  of  any  number  between  3  and  +  oc. 

Hence  the  given  equation  has  3  real  roots  j  one  situated  be 
tween  —2  and  —3,  one  between  2  and  2-J-,  and  a  third  between 
2j-  and  3.  The  initial  figures  of  the  roots  are  therefore  —2, 
+  2,  and  +2. 

There  are  three  changes  of  sign  of  the  primitive  function, 
two  of  the  first  derived  function,  and  one  of  the  second  derived 
function  ;  but  no  variation  is  lost  by  the  change  of  sign  of 
either  of  the  derived  functions;  while  every  change  of  sign  of 
the  primitive  function  occasions  a  loss  of  one  variation. 
•  2.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x3—  5x2  +  8x—  1  =  0. 

Here  we  have 


^=-2295. 

When  x=  —  cc,  the  signs  are  —  -\ ,  giving  2  variations, 

x=+(xj  il  +  +  H "1  " 

Hence  this  equation  has  but  one  real  root,  and,  consequently, 
must  have  two  imaginary  roots.  Moreover,  it  is  easily  proved 
that  the  real  root  lies  between  0  and  +1. 

3.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  ^4  °^3 

Here  we  have 


,  or  2x3-3x2-7x+5, 
K=17x2-23x-45, 


R2=+ 524535. 
When  x=  —  oc,  the  signs  are  +  —  +  —  +,  giving  4  variations, 

Hence  the  four  roots  of  this  equation  are  real. 


33  0  ALGEBRA. 

Substituting  different  values  for  x,  we  find  that 

when  x=  —  3,  the  signs  are  H  ----  1  ----  h,  giving  4  variations, 

x=-2,  _  +  +-  +  ,       »      3 

05=  -1,  _  +  +  __+,       «      3 

a?=     0,  +  +  -     -  +  ,       "      2         " 

#=  +  1,  "           +  -    --+,       "      2 

x=+2,  "           H  -----  h,       "      2         " 


a=+3,  +  +  +  +  +,       "      0 

Hence  this  equation  has  one  negative  root  between  —2  and 
—3,  one  negative  root  between  0  and  —  1,  one  positive  root 
between  2  and  2-J,  and  another  positive  root  between  2J  and  3. 

4.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x3—  7x4-7  =  0. 

Ans.  Three;  viz.,  one  between  —3  and.  —4,  one  be 
tween  1  and  1-J,  and  the  other  between  1-J  and  2. 

5.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  2x4  —  20x  +  19  =  0. 

Ans.  Two;  viz.,  one  between  1  and  1-J,  the  other 
between  \\  and  2. 

6.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x5  +  2x4  +  3^3  +  4x2  +  5x—  20  =  0. 

Ans.  One,  situated  between  1  and  2. 

7.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x3  +  3x2  +  5x—  178  =  0. 

Ans.  One,  situated  between  4  and  5. 

8.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x4-12x2  +  12x-3  =  0. 

Ans.  Four;  viz.,  one  between  —3  and  —4,  one  be 
tween  0  and  -J,  one  between  %  and  1,  and  the 
other  between  2  and  3. 

9.  Find  the  number  and  situation  of  the  real  roots  of  the 
equation  x4-8x3  +  14x2  +  4x-8  =  0. 

Ans.  Four;  viz.,  one  between  —1  and  0,  one  between 
0  and  +1,  one  between  2  and  3,  and  the  other 
between  5  and  6. 


GENERAL  THEORY  OF  EQUATIONS.          331 

Solution  of  Simultaneous  Equations  of  any  Degree. 

459.  One  of  the  most  general  methods  for  the  elimination 
of  unknown  quantities  from  a  system  of  equations,  depends 
upon  the  principle  of  the  greatest  common  divisor. 

Suppose  we  have  two  equations  involving  x  and  y.  We 
first  transpose  all  the  terms  to  one  member,  so  that  the  equa 
tions  will  be  of  the  form 

A=0,  B  =  0. 

We  arrange  the  terms  in  the  order  of  the  powers  of  a?,  and 
we  will  suppose  that  the  polynomial  B  is  not  of  a  higher  de 
gree  than  A.  We  divide  A  by  B,  as  in  the  method  of  finding 
the  greatest  common  divisor,  Art.  95,  and  continue  the  opera 
tion  as  far  as  possible  without  introducing  fractional  quotients 
having  x  in  the  denominator.  Let  Q  represent  the  quotient, 
and  E  the  remainder  ;  we  shall  then  have 


But,  since  A  and  B  are  each  equal  to  zero,  it  follows  that  R 
must  be  equal  to  zero.  If,  then,  there  are  certain  values  of  x 
and  y  which  render  A  and  B  equal  to  zero,  these  values  should 
be  the  roots  of  the  equations 

B  =  0,  R  =  0. 

We  now  divide  B  by  K,  and  continue  the  operation  as  far  as 
possible  without  introducing  fractional  quotients  having  x  in 
the  denominator.  Let  R/  denote  the  remainder  after  this  divi 
sion.  For  the  same  reason  as  before,  R'  must  equal  zero,  and 
we  thus  obtain  the  two  equations 

R  =  0,  R'  =  0, 

whose  roots  must  satisfy  the  equations  A  =  0,  B  —  0.  If  we  con 
tinue  to  divide  each  remainder  by  the  succeeding,  and  suppose 
that  each  remainder  is  of  a  lower  degree  with  respect  to  x  than 
the  divisor,  we  shall  at  last  obtain  a  remainder  which  does 
not  contain  x.  Let  R"  denote  this  remainder.  The  equation 
R7/  =  0  will  furnish  the  values  of  ?/,  and  the  equation  R'  =  0  will 
furnish  the  corresponding  values  of  x. 

If  we  have  three  equations  involving  three  unknown  quan 
tities,  we  commence  by  reducing  them  to  two  equations  with 


332  ALGEBRA. 

two  unknown  quantities,  and  subsequently  to  a  single  final 
equation  by  a  process  similar  to  that  above  explained. 

f      2    I        2        ~l  Q        i\ 

Ex.  1.  Solve  the  two  equations  j  ~ 

Divide  the  first  polynomial  by  the  second,  as  follows : 
x  +  y— 5 


-(y-6)ag- 


2?/2—  1%+12,  the  remainder. 
This  remainder  must  be  equal  to  zero  ;  that  is, 


whence  2/  —  2  or  3. 

When  y=2,  x  =  3; 


Ex.  2.  Solve  the  equations  |  ' 

[xy+xy2—  12  =  0. 

Multiply  the  first  polynomial  by  y,  to  make  its  first  term  di 
visible,  and  proceed  as  follows  : 


12—  3  0#  -hi  2  ?/2,  the  remainder. 

Hence  12-30?/  +  12?/2  =  0; 

therefore  2/=z2  or  -J-. 

When  =  2x  =  2 


Ex.  3.  Solve  the  equations  |  ^y- 

(  cc3?/  —  x3  +  a:—  3  =  0. 


The  first  remainder  is  3?/  —  3,  which,  being  placed  equal  to 
0,  gives  y  =  l,  whence  a?  =3. 
Ex.  4.  Solve  the  equations 

(3?/2—  ?y  +  l)-?/3  +  ?/2—  2?/  =  0, 


The  remainder  after  the  first  division  is  x—  2y,  and  after  the 
second  division  y2  —  y.     Hence  we  conclude 
x  —  2y  =  0,  and  y2—y  —  0. 


GENERAL  THEORY  OF  EQUATIONS.         833 

Whence  we  have  y=l  or  0, 

x  =  2  or  0. 


v  -  r   o.i     *!.  *• 

Ex.5.  Solve  the  equations 

The  remainder  after  the  first  division  is 


Hence  we  have  12  (y—  l)(a?—  1)  =  0,  which  equation  may  be 
satisfied  by  supposing  y—  1  =  0  or  x—  1  =  0. 
When  a?  =  l,  y——  1  or  0, 

y  =  l,  ar=     2  or  3. 

IT     fl   G  i      *i  r 

Ex.6.  Solve  the  equations 


The  first  division  gives  a  remainder  cc(y—  2)+2/2—  4,  whence 
we  have 


^ 
and  we  may  have  either 

y—  2  =  0,  or  ic 

If  we  divide  the  first  member  of  the  second  equation  by 
jc+^-{-2,  we  obtain  the  remainder  ?/2—  5?/+6,  which  also  equals 
zero;  whence  y=2  or  3. 

When  y  =  2,  x=  —4  or  0, 

y=3,  o;=-5. 


334  ALGEBKA. 


CHAPTEE  XXII. 

SOLUTION   OF  NUMERICAL  EQUATIONS  OF  HIGHER  DEGREES. 

461.  Equations  of  the  third  and  fourth  degrees  can  some« 
times  be  solved  by  direct  methods ;  but  these  methods  are  com 
plicated,  and  are  of  limited  application.     No  general  solution 
of  an  equation  higher  than  the  fourth  degree  has  yet  been  dis 
covered.    To  obtain  the  roots  of  numerical  equations  of  degrees 
higher  than  the  second,  we  must  generally  employ  tentative 
methods,  or  methods  which  involve  approximation. 

462.  Commensurable  Roots  of  an  Equation. — Any  equation 
having  fractional  coefficients  can  be  transformed  into  another 
which  has  all  its  coefficients  integers,  and  the  coefficient  of  its 
first  term  unity,  Art.  441,  and  such  an  equation  can  not  have 
a  root  which  is  a  rational  fraction,  Art.  440 ;  that  is,  every  com 
mensurable  root  of  this  equation  must  be  an  integer.     Every 
integral  root  of  this  equation  is  a  divisor  of  the  last  term,  Art. 
439.     Hence,  to  find  the  commensurable  roots  of  an  equation, 
we  need  only  make  trial  of  the  integral  divisors  of  the  last  term. 

463.  Method  of  finding  the  Roots. — In  order  to  discover  a  con 
venient  method  of  finding  the  roots,  we  will  form  the  equation 
whose  roots  are  2,  3,  4,  and  5.     This  equation,  Art.  436,  may 
be  expressed  thus, 

(a- 2)  (a- 3)  (a- 4)  (a- 5)  =  0. 

If  we  perform  the  multiplication  here  indicated,  we  shall  ob 
tain  £4-14x3  +  71x2-154;r+120  =  0. 

"We  know  that  this  equation  is  divisible  by  x— 5,  and  we 
will  perform  the  operation  by  an  abridged  method.  Since  the 
coefficients  of  the  quotient  depend  simply  upon  the  coefficients 
of  the  divisor  and  dividend,  and  not  upon  the  literal  parts  of 
the  terms,  we  may  obtain  the  coefficients  of  the  quotient  by  op 
erating  upon  the  coefficients  of  the  divisor  and  dividend  by  the 


NUMERICAL   EQUATIONS  OF  HIGHER  DEGREES.  335 

usual  method.  To  the  coefficients  thus  found  the  proper  let 
ters  may  afterward  be  annexed.  The  operation  may  then  be 
exhibited  as  follows : 

A     B      C       D       ^  r 


1-14  +  71-154+120 
1-  5 


1—5,  divisor, 

1  —  9  +  26—24,  quotient. 


-  9  +  71 

-  9+45 


+  26-154 
+  26-130 


-  24  +  120 
-  24  +  120. 

Supplying  the  powers  of  x,  we  obtain  for  a  quotient 


In  applying  this  method  of  division,  care  should  be  taken  to 
arrange  the  terms  in  the  order  of  the  powers  of  x  ;  and  if  the 
series  of  powers  of  x  in  the  dividend  is  incomplete,  we  must 
supply  the  place  of  the  deficient  term  by  a  cipher. 

The  preceding  operation  may  be  still  further  abridged  by 
performing  the  successive  subtractions  mentally,  and  simply 
writing  the  results.  Eepresent  the  root  5  by  r,  and  the  coeffi 
cients  of  the  given  equation  by  A,  B,  C,  D,  .  .  .  .  V. 

We  first  multiply  —  r  by  A,  and  subtract  the  product  from 
B;  the  remainder,  —  9,  we  multiply  by  —  r,  and  subtract  the 
product  from  C;  the  remainder,  +26,  we  multiply  by  —r,  and 
subtract  the  product  from  D  ;  the  remainder,  —24,  we  multi 
ply  by  —  r,  and,  subtracting  from  V,  nothing  remains.  If  we 
take  the  root  r  with  a  positive  sign,  we  may  substitute  in  the 
above  process  addition  for  subtraction  ;  and  if  we  set  down  only 
the  successive  remainders,  the  work  will  be  as  follows  : 

A    B      C      D       V     r 

1-14+71-154+120(5 

1_  9  +  26-  24, 
and  the  rule  will  be 

Multiply  A  by  r,  and  add  the  product  to  B  ;  set  down  the  sum, 
multiply  it  ~by  r,  and  add  the  product  to  C  ;  set  down  the  sum,  mul- 


336  ALGEBRA. 

tiply  it  by  r,  and  add  the  product  to  D,  and  so  on.  The  final  prod 
uct  should  be  equal  to  the  last  term  V,  taken  with  a  contrary  sign. 
The  coefficients  above  obtained  are  the  coefficients  of  a  cubic 
equation  whose  roots  are  2,  3,  and  4.  The  polynomial  may 
therefore  be  divided  by  x  —  4,  and  the  operation  will  be  as 
follows  : 


1-5+  6. 

These,  again,  are  the  coefficients  of  a  quadratic  equation 
whose  roots  are  2  and  3.     Dividing  again  by  x—  3,  we  have 

1-5  +  6(3 
1-2, 
which  are  the  coefficients  of  the  binomial  factor  x  —  2. 

These  three  operations  of  division  may  be  exhibited  together 
as  follows: 


1-14+71-154  +  120 
1_  9  +  26-  24 
1-  5+   6 
1-  2 


5,  first  divisor. 
4,  second  divisor. 
3,  third  divisor. 


464.  How  to  find  all  the  Integral  Roots.  —  The  method  here  ex 
plained  will  enable  us  to  find  all  the  integral  roots  of  an  equa 
tion.  For  this  purpose,  we  make  trial  of  different  numbers  in 
succession,  all  of  which  must  be  divisors  of  the  last  term  of  the 
equation.  If  any  division  leaves  a  remainder,  we  reject  this  di 
visor  ;  if  the  division  leaves  no  remainder,  the  divisor  employed 
is  a  root  of  the  equation.  Thus,  by  a  few  trials,  all  the  integral 
roots  may  be  easily  found. 

The  labor  will  often  be  diminished  by  first  finding  positive 
and  negative  limits  of  the  roots,  for  no  number  need  be  tried 
which  does  not  fall  within  these  limits. 

Ex.  2.  Find  the  seven  roots  of  the  equation 


We  take  the  coefficients  separately,  as  in  the  last  example, 
and  try  in  succession  all  the  divisors  of  36,  both  positive  and 
negative,  rejecting  such  as  leave  a  remainder.  The  operation 
is  as  follows: 


NUMERICAL   EQUATIONS  OF  HIGHER  DEGREES. 


337 


1  +  1-14-14+49+49-36-36 

14-2-12-26  +  23  + 

l  +  4_  4-34-45-18 

1  +  74.17  +  17+  6 

1  +  6  +  11+  6 

1+5+  6 

1  +  3 
Hence  the  seven  roots  are 

1,  2,  3,  -1,  -1,  -2,  -3. 
Ex.  3.  Find  the  six  roots  of  the  equation 


1,  first  divisor. 

2,  second  divisor. 

3,  third  divisor. 
— 1,  fourth  divisor. 
—  1,  fifth  divisor. 
—2,  sixth  divisor. 
—3,  seventh  divisor, 


—  1584=0. 


1+  5-81-  85  +  964+  780-1584 

1+  6-75-160  +  804  +  1584 

1  +  10-35-300-396 

1  +  16  +  61+  66 

1  +  14+33 

1  +  11 
The  six  roots,  therefore,  are 

1,  4,  6,  -2,  -3,  -11. 
Ex.  4.  Find  the  five  roots  of  the  equation 


1. 

4. 

6. 

-  2. 

-  3. 

-IL 


1  +  6-10-112-207-110 


1. 


1  +  5-15-  97-110  -2. 

1  +  3-21-  55  -5. 

1-2-11 
Three  of  the  roots,  therefore,  are 

-1,  -2,  -5. 

The  two  remainining  roots  may  be  found  by  the  ordinary 
method  of  quadratic  equations.  Supplying  the  letters  to  the 
last  coefficients,  we  have 

xz-2x-ll  =  0. 

Hence  cc~l±VT2. 

Ex.  5.  Find  the  four  roots  of  the  equation 


Ans.  1,  2,  3,  and  6. 

Ex-  6.  Find  the  four  roots  of  the  equation 


338  ALGEBRA. 

Ex.  ?t,  Find  the  four  roots  of  tbe  equation 
x4_55x2-30z+  504  =  0. 
Ex.  8.  Find  the  four  roots  of  the  equation 


Ex.  9.  Find  the  four  roots  of  the  equation 
x4-z3-z2-fl9z-42  =  0. 
Ex.  10.  Find  the  five  roots  of  the  equation 


465.  Incommensurable  Roots.  —  If  a  high  numerical  equation 
is  found  to  contain  no  commensurable  roots,  or,  if  after  remov 
ing  the  commensurable  roots,  the  depressed  equation  is  still 
of  a  higher  degree  than  the  second,  we  must  proceed  by  ap 
proximation  to  find  the  incommensurable  roots.  Different 
methods  may  be  employed  for  this  purpose  ;  but  the  following 
method,  which  is  substantially  the  same  as  published  by  Horner 
in  1819,  is  generally  to  be  preferred. 

Find,  by  Sturm's  Theorem,  or  by  trial,  Art.  446,  the  integral 
part  of  a  root,  and  transform  the  given  equation  into  another 
whose  roots  shall  be  less  than  those  of  the  preceding  by  the 
number  just  found,  Art.  444.  Find,  by  Art.  446,  the  first  fig 
ure  of  the  root  of  this  equation,  which  will  be  the  first  decimal 
figure  of  the  root  of  the  original  equation.  Transform  the  last 
equation  into  another  whose  roots  shall  be  less  than  those  of 
the  preceding  by  the  figure  last  found.  Find,  as  before,  the 
first  figure  of  the  root  of  this  equation,  which  will  be  the  sec 
ond  decimal  figure  of  the  root  of  the  original  equation.  By 
proceeding  in  this  manner  from  one  transformation  to  another, 
we  may  discover  the  successive  figures  of  the  root,  and  may 
carry  the  approximation  to  any  degree  of  accuracy  required. 

Ex.  1.  Find  an  approximate  root  of  the  equation 


We  have  found,  page  330,  that  this  equation  has  but  one  real 
root,  and  that  it  lies  between  4  and  5.  The  first  figure  of  the 
root  therefore  is  4.  Transform  this  equation  into  anotner  whose 
roots  shall  be  less  than  those  of  the  proposed  equation  by  4, 
which  is  done  by  substituting  2/4-4  for  x.  "We  thus  obtain 

' 


NUMERICAL  EQUATIONS  OF    HIGHER  DEGREES.          339 

The  first  figure  of  the  root  of  this  equation  is  .5.     Transform 
the  last  equation  into  another  whose  roots  shall  be  less  by  .5, 
which  is  done  by  substituting  z-f.5  for  y.     We  thus  obtain 
z3-fl6.5224-92.75z=3.625. 

The  first  figure  of  the  root  of  this  equation  is  .03.  Transform 
the  last  equation  into  another  whose  roots  shall  be  less  by  .03, 
which  is  done  by  substituting  v-\-.OB  for  z.  We  thus  obtain 


The  first  figure  of  the  root  of  this  equation  is  .008.  Trans 
form  the  last  equation  into  another  whose  roots  shall  be  less 
by  .008,  and  thus  proceed  for  any  number  of  figures  required. 

486.  How  the  Operation  may  be  abridged.  —  This  method  would 
be  very  tedious  if  we  were  obliged  to  deduce  the  successive 
equations  from  each  other  by  the  ordinary  method  of  substitu 
tion  ;  but  they  may  be  derived  from  each  other  by  a  simple 
law.  Thus,  let 

Ax34-Bx2+Cx=V  (1.) 

be  any  cubic  equation,  and  let  the  first  figure  of  its  root  be  de 
noted  by  r,  the  second  by  r',  the  third  by  r",  and  so  on. 

If  we  substitute  r  for  x  in  equation  (1),  we  shall  have 
Ar3  +  Br2-f  O  =  V,  nearly, 

Whence  r=~     _T     .    ,.  (2.) 

C  +  Br-f  A?-2 

If  we  put  y  for  the  sum  of  all  the  figures  of  the  root  except 
the  first,  we  shall  have  x=r-\-y  ;  and,  substituting  this  value 
for  x  in  equation  (1),  we  obtain 


4-   Br2+2Bn/  -fB?/2  V  =V; 

+  Cr       +  Cy    j 
or,  arranging  according  to  the  powers  of  y,  we  have 


Let  us  put  Br  for  the  coefficient  of  y2,  C'  for  the  coefficient  of 
?/,  and  V7  for  the  right  member  of  the  equation,  and  we  have 
A^+By  +  C'y=V.  (3.) 

This  equation  is  of  the  same  form  as  equation  (1)  ;  and,  pro 
ceeding  in  the  same  manner,  we  shall  find 


840  ALGEBRA. 

V7 


where  r'  is  the  first  figure  of  the  root  of  equation  (3),  or  the 
second  figure  of  the  root  of  equation  (1). 

Putting  z  for  the  sum  of  all  the  remaining  figures,  we  have 
y=r'  +  z;  and,  substituting  this  value  in  equation  (3),  we  shall 
obtain  a  new  equation  of  the  same  form,  which  may  be  written 


and  in  the  same  manner  we  may  proceed  with  the  remaining 
figures. 

Equation  (2)  furnishes  the  value  of  the  first  figure  of  the  root  ; 
equation  (4)  the  second  figure,  and  similar  equations  would  fur 
nish  the  remaining  figures.  Each  of  these  expressions  involves 
the  unknown  quantity  which  is  sought,  and  might  therefore 
appear  to  be  useless  in  practice.  When,  however,  the  root  has 
been  found  to  several  decimal  places,  the  value  of  the  terms 

Br  and  Ar2  will  be  very  small  compared  with  C,  and  r  will  be 

y 
very  nearly  equal  to  —  .     We  may  therefore  employ  C  as  an 

0 

approximate  divisor,  which  will  probably  furnish  a  new  figure 
of  the  root.  Thus,  in  the  last  example,  all  the  figures  of  the 
root  after  the  first  are  found  by  division. 

46-77      =.5, 
3.62  -92.75  =  .03, 


If  we  multiply  the  first  coefficient  A  by  r,  the  first  figure 
of  the  root,  and  add  the  product  to  the  second  coefficient,  we 
shall  have 

B  +  Ar.  (6.) 

If  we  multiply  expression  (6)  by  r,  and  add  the  product  to 
the  third  coefficient,  we  shall  have 

C  +  Br+Ar2.  (7.) 

If  we  multiply  expression  (7)  by  r,  and  subtract  the  product 
from  V,  we  shall  have 

Y_O-Br2-Ar3, 
which  is  the  quantity  represented  by  Vx  in  equation  (3). 


NUMERICAL   EQUATIONS   OF   HIGHER   DEGREES.  341 

If  we  multiply  the  first  coefficient  A  by  r,  and  add  the  prod 
uct  to  expression  (6),  we  shall  have 

B  +  2Ar.  (8.) 

If  we  multiply  expression  (8)  by  r,  and  add  the  product  to 
expression  (7),  we  shall  have 

C  +  2Br+3Ar2, 
which  is  the  coefficient  of  y  in  equation  (3). 

If  we  multiply  the  first  coefficient  A  by  r,  and  add  the  prod 
uct  to  expression  (8),  we  shall  have 


which  is  the  coefficient  of  y2  in  equation  (3). 

We  have  thus  obtained  the  coefficients  of  the  first  transformed 
equation ;  and,  by  operating  in  the  same  manner  upon  these 
coefficients,  we  shall  obtain  the  coefficients  of  the  second  trans 
formed  equation,  and  so  on ;  and  the  successive  figures  of  the 
root  are  indicated  by  dividing  V  by  C,  Y'  by  C',  Y"  by  C",  etc. 

467.  The  results  of  the  preceding  discussion  are  expressed 
in  the  following 

RULE. 

Represent  the  coefficients  of  the  different  terms  by  A,  B,  C,  and 
the  right-hand  member  of  the  equation  by  Y.  Having  found  r,  the 
first  figure  of  the  root,  multiply  A  by  r,  and  add  the  product  to  B. 
Set  down  the  sum  under  B ;  multiply  this  sum  by  r,  and  add  the 
product  to  C.  Set  down  the  sum  under  C ;  multiply  it  by  r,  and 
subtract  the  product  from  Y;  the  remainder  will  be  the  FIRST  DIV 
IDEND. 

Again,  multiply  A  by  r,  and  add  the  product  to  the  last  number 
under  B.  Multiply  this  sum  by  r,  and  add  the  product  to  the  last 
number  under  C ;  this  result  will  be  the  FIRST  TRIAL  DIVISOR. 

Again,  multiply  A  by  r,  and  add  the  product  to  the  last  number 
under  B. 

Find  the  second  figure  of  the  root  by  dividing  the  first  dividend 
by  the  first  trial  divisor,  and  proceed  with  this  second  figure  pre 
cisely  as  was  done  with  the  first  figure,  carefully  regarding  the  local 
value  of  the  figures. 

The  second  figure  of  the  root  obtained  by  division  will  fre- 


342 


ALGEBRA. 


qently  furnish  a  result  too  large  to  be  subtracted  from  the  re 
mainder  V7,  in  which  case  we  must  assume  a  different  figure. 
After  the  second  figure  of  the  root  has  been  obtained,  there  will 
seldom  be  any  further  uncertainty  of  this  kind. 

It  may  happen  that  one  of  the  trial  divisors  becomes  zero. 
In  this  case  equation  (2)  becomes 


whence 


=  ™   Or  r  —  \/  -=: 


that  is,  the  next  figure  of  the  root  will  be  indicated  by  dividing 
the  last  dividend  by  the  last  number  under  B,  and  extracting 
the  square  root  of  the  quotient. 

The  entire  operation  for  finding  a  root  of  the  equation 


may  be  exhibited  as  follows  : 


A     B 

1     +3 
4 
7 
4 
11 
4 

C                         Y       r 

+5                   =178    (4.5388=x. 
28                       132 

33                          46=  1st  dividend. 
44                         42.375 

77  =  1st  divisor. 

7.75 

3.625  =2d  dividend. 
2.797377 

15.5 
.5 

84.75 
8.00 
92.75  =2d  divisor. 
.4959 

.827623  =  3d  dividend. 
.751003872 

16.0 
.5 

.076619128=  4th  dividend 

16.53 
3 
16.56 
3 

93.2459 
.4968 

93.7427  =  3d  divisor. 
.132784 

16.598 
8 

93.875484 
.132848 

16.606     94.008332 =4th  divisor. 
Having  found  one  root,  we  may  depress  the  equation 


NUMERICAL  EQUATIONS  OF  HIGHER  DEGREES.          343 

to  a  quadratic  by  dividing  it  by  x—  4.5388.     We  thus  obtain 


where  x  is  evidently  imaginary,  because  q  is  negative  and 
greater  than  £     See  Art.  280. 

After  thus  obtaining  the  root  to  five  or  six  decimal  places, 
several  more  figures  will  be  correctly  obtained  by  simply  di 
viding  the  last  dividend  by  the  last  divisor. 

Ex.  2.  Find  all  the  roots  of  the  equation 


The  first  figure  of  one  of  the  roots  we  readily  find  to  be  3. 
We  then  proceed,  according  to  the  Rule,  to  obtain  the  root  to 
four  decimal  places,  after  which  two  more  will  be  obtained 
correctly  by  division. 
ABC  Y         r 

I    +  11  -102  =-181      (3.21312=x. 

_3        42  _180 

14—60  -1  =  1st  dividend. 

_3        51  -.992 

17       -^9  =  1st  divisor.        -  .008"  =  2d  dividend. 
3          4.04  -.006739 

20.2    -4.96  -.001261  =  3d  dividend. 

_2       4.08  -.001217403 

20.4    -  0.88  =  2d  divisor.    -.000043597  =  4th  dividend. 
2          .2061 

20.61  -.6739 
_  1        .2062 

20.62  -.4677  =  3d  divisor. 
_  1_      .061899 

2O633  -.405801 
_  8  .061908 
20.636  —  .343893  =  4th  divisor. 

The  two  remaining  roots  may  be  found  in  the  same  way,  or 
by  depressing  the  original  equation  to  a  quadratic.     Those 

roots  are, 

3.22952 
—17.44265. 


344  ALGEBKA. 

When  a  power  of  x  is  wanting  in  the  proposed  equation,  we 
must  supply  its  place  with  a  cipher. 

Ex.  3.  Find  all  the  roots  of  the  cubic  equation 
33-70;=:-.  7. 

The  work  of  the  following  example  is  exhibited  in  an  ab 
breviated  form.  Thus,  when  we  multiply  A  by  r,  and  add  the 
product  to  B,  we  set  down  simply  this  result.  We  do  the  same 
in  the  next  column,  thus  dispensing  with  half  the  number  of 
lines  employed  in  the  preceding  example.  Moreover,  we  may 
omit  the  ciphers  on  the  left  of  the  successive  dividends,  if  we 
pay  proper  attention  to  the  local  value  of  the  figures.  Thus 
it  will  be  seen  that  in  the  operation  for  finding  each  successive 
figure  of  the  root,  the  decimals  under  B  increase  one  place, 
those  under  C  increase  two  places,  and  those  under  Y  increase 
three  places. 
1  +  0  .-7  =  -7  (1.356895867=*. 

1  -6  -6 

2  —4=  1st  div'r.      —  1=  1st  dividend. 
3.3          -3.01  -.903 

3.6          -1.93=2d  div'r.     -97=  2d  dividend. 
3.95        -1.7325  86625 

4.00        r.  1.5325=  3d  div'r.     10375=  3d  dividend. 
4.056      -1.508164  9048984 

4.062      -1.483792  =4th  div'r.  1326016=  4th  dividend. 
4.0688    -1.48053696  1184429568 

4.0696    -1.47728128  =  5th  div.  141586432  =  5th  div'd. 
4.07049  -1.4769149359  132922344231 

4.07058  -  1,4765485837  =  6th  div.  8664087769  =  6th  div'd 
Having  proceeded  thus  far,  four  more  figures  of  the  root, 

5867,  are  found  by  dividing  the  sixth  dividend  by  the  sixth  di 

visor. 

We  may  find  the  two  remaining  roots  by  the  same  process  ; 

or,  after  having  obtained  one  root,  we  may  depress  the  equation 


to  a  quadratic  equation  by  dividing  by  #—1.356895867,  and 
we  shall  obtain 

x2  4-  1.35689586705-5.158833606  =  0. 


NUMERICAL   EQUATIONS  OF   HIGHER   DEGREES.          345 

Solving  this  equation,  we  obtain  _ 
x=-.  678447933  ±V5.  619125204. 

(  -3.048917, 

Hence  the  three  roots  are    ......    <      1.356896, 

^r  (      1.692021. 

Ex.  4.  Find  a  root  of  the  equation  2#3  +  3x2  =  850. 
2+3  +0  =850     (7.0502562208 

17  119  833 

31  336  =  1st  divisor.       17  =  1st  dividend. 

45.10       338.2550  16.912750 

45.20       340.5150=  2d  divisor.     87250=2d  dividend. 

45.3004   340.52406008  68104812016 

45.3008    340.533  12024=  3d  div.  19145187984=  3d  div'd. 

45.30130  340.5353853050  17026769265250 

45.30140  340.5376503750  =4th  d.    2118418718750=4th  div. 

Dividing  the  fourth  dividend  by  the  fourth  divisor,  we  ob 
tain  the  figures  62208,  which  make  the  root  correct  to  the 
tenth  decimal  place. 

The  two  remaining  values  of  x  may  be  easily  shown  to  be 
imaginary. 

When  a  negative  root  is  to  be  found,  we  change  the  signs 
of  the  alternate  terms  of  the  equation,  Art.  442,  and  proceed 
as  for  a  positive  root. 

Ex.  5.  Find  a  root  of  the  equation  5x3—  6a?2  +  3ic=—  85. 

Changing  the  signs  of  the  alternate  terms,  it  becomes 


+6          +3  +85     (2.16139. 

16  35  70 

26  87  =  1st  divisor.          15  =  1st  dividend. 

36.5        90.65  9.065 

37.0         94.35  =  2d  divisor.       5.935  =  2d  dividend. 
37.80      96.6180  6.797080 

.38.10      98.9040=  3d  divisor.      137920=  3d  dividend. 
38.405     98.942405  98942405 

38.410     98.980815  =  4th  divisor.  38977595  =  4th  div'd. 
38.4165  98.99233995  29697701985 

38.4180  99.00386535=  5th  div'r.     9279893015  =5th  div'd. 

P  2 


346  ALGEBRA. 

Hence  one  root  of  the  equation 


=—  85 
is  -2.16139. 

The  same  method  is  applicable  to  the  extraction  of  the  cube 
root  of  numbers. 

7':  Ex.  6.  Let  it  be  required  to  extract  the  cube  root  of  9  ;  in 
Dther  words,  it  is  required  to  find  a  root  of  the  equation 

#3  =  9. 

1+0  +0  =9     (2.0800838. 

24  8 

4  12  =  1st  divisor.      1  =  1st  dividend. 

6.08          12.4864  .998912 

6.16         12.9792  =  2d  divisor.     1088  =  2d  dividend. 
6.24008    12.9796992064  1038375936512 

6.24016    12.9801984192  =  3dd.      49624063488=  3d  div. 
6.240243  12.980217139929  38940651419787 

6.240246  12.980235860667  =4thd.  10683412068213  =4thd. 

Ex.  7.  Find  all  the  roots  of  the  equation 


(  1.02804. 
Ans.  \  6.57653. 
(  7.39543. 
Ex.  8.  Find  all  the  roots  of  the  equation 


(  -1.12061. 
Ans.  •]  —3.34730. 
(  -4.53209. 
Ex.  9.  Extract  the  cube  root  of  48228544. 

Ans.   364. 

Ex.  10.  There  are  two  numbers  whose  difference  is  2,  and 
whose  product,  multiplied  by  their  sum,  makes  100.  What 
are  those  numbers  ? 

Ex.  11.  Find  two  numbers  whose  difference  is  6,  and  such 
that  their  sum,  multiplied  by  the  difference  of  their  cubes,  may 
produce  5000. 

Ex.  12.  There  are  two  numbers  whose  difference  is  4  ;  and 


NUMERICAL   EQUATIONS  OF  HIGHER  DEGREES.          347 

the  product  of  this  difference,  by  the  sum  of  their  cubes,  is 
3400.     What  are  the  numbers  ? 

Ex.  13.  Several  persons  form  a  partnership,  and  establish  a 
certain  capital,  to  which  each  contributes  ten  times  as  many 
dollars  as  there  are  persons  in  company.  They  gain  6  plus 
the  number  of  partners  per  cent.,  and  the  whole  profit  is  $392. 
How  many  partners  were  there  ? 

Ex.  14.  There  is  a  number  consisting  of  three  digits  such 
that  the  sum  of  the  first  and  second  is  9  ;  the  sum  of  the  first 
and  third  is  12  ;  and  if  the  product  of  the  three  digits  be  in 
creased  by  38  times  the  first  digit,  the  sum  will  be  336.  Ke= 
quired  the  number. 

(      636, 

Ans.  1  or  725, 

(  or  814. 

Ex.  15.  A  company  of  merchants  have  a  common  stock  of 
$4775,  and  each  contributes  to  it  twenty-five  times  as  many 
dollars  as  there  are  partners,  with  which  they  gain  as  much 
per  cent,  as  there  are  partners.  Now,  on  dividing  the  profit, 
it  is  found,  after  each  has  received  six  times  as  many  dollars 
as  there  are  persons  in  the  company,  that  there  still  remains 
$126.  Kequired  the  number  of  merchants. 

^  Ans.  7,  8,  or  9. 

EQUATIONS  OF  THE   FOURTH  AND  HIGHER  DEGREES. 

468.  It  may  be  easily  shown  that  the  method  here  employed 
for  cubic  equations  is  applicable  to  equations  of  every  degree. 
For  the  fourth  degree  we  shall  have  one  more  column  of  prod 
ucts,  but  the  operations  are  all  conducted  in  the  same  manner, 
as  will  be  seen  from  the  following  example. 

Ex.  1.  Find  the  four  roots  of  the  equation 


By  Sturm's  Theorem,  we  have  found  that  these  roots  are  all 
real;  three  positive,  and  one  negative. 
We  then  proceed  as  follows  : 


348 


ALGEBRA. 


1-8 
-3 
+  2 
7 
12.2 
12.4 
12.6 
12.83 
12.86 
12.89 

+  14        +4                        =8     (5.2360679. 
-   1        -   I                         -5 
-f   9        +  44  =  1st  divisor.     13  =  1st  dividend. 
44           53.288                   10.6576 

46.44      63.072  =  2d  div'r.    2.3424  =2d  dividend. 
48.92       64.626747               1.93880241 

51.44       66.193068  =  3ddiv.  .40359759  =  3d  div'd. 
51.8249  66.509117736            .399054706416 

52.2107  66.825633024=4thd.    4542883584=  4th  d 

52.5974 

12.926  52.674956 
12.932  52.752548 
and  by  division  we  obtain  the  four  figures  0679. 

The  other  three  roots  may  be  found  in  the  same  manner. 

C-  .7320508, 


Hence  the  four  roots  are 


5.2360679. 
Ex.  2.  Find  a  root  of  the  equation 

x5  +  2^4  +  3x3  +  4x2  +  5x  =  20. 

"We  have  found,  by  Sturm's  Theorem,  that  this  equation  has 
a  real  root  between  1  and  2. 
"We  then  proceed  as  follows  : 


1+2 

+  3 

+4 

+5 

+20   (1.125789. 

3 

6 

10 

15 

15 

4 

10 

20 

35=  1st  divisor. 

~5  =  1st  dividend. 

5 

15 

35 

38.7171 

3.87171 

6 

21 

37.171 

42.6585  =2d  divisor. 

1.12829  =2d  dividend. 

7.1 

21.71 

39.414 

43.5027 

2016 

.87005 

44032 

7.2 

22.43 

41.730 

44.3566 

2080,  3d  div'r. 

.25823 

55968,  3d  div'd. 

7.3 

23.16 

42.211 

008 

44.5731 

44750625 

.22286 

5723753125 

7.4 

23.90 

42.695 

032 

44.7902 

83203125,  4th 

d.   3536 

9873046875, 

4th  d. 

7.5 

2 

24.05 

04 

43.182 

080 

7.5 

4 

24.20 

12 

43.304 

790125 

7.5 

f> 

24.35 

24 

43.427 

690500 

7.5 

8 

24.50 

40 

7.6 

0,5 

24.54 

2025 

7.6 

10 

24.58 

0075 

Dividing  the  fourth  dividend  by  the  fourth  divisor,  we  ob 
tain  the  figures  789, 


NUMERICAL   EQUATIONS  OF   HIGHER   DEGREES.          349 

When  we  wish  to  obtain  a  root  correct  to  a  limited  number 
of  places,  we  may  save  much  of  the  labor  of  the  operation  by 
cutting  off  all  figures  beyond  a  certain  decimal.  Thus  if,  in  the 
example  above,  we  cut  off  all  beyond  five  decimal  places  in  the 
successive  dividends,  and  all  beyond  four  decimal  places  in  the 
divisors,  it  will  not  affect  the  first  six  decimal  places  in  the  root. 
Ex.  3.  Find  the  roots  of  the  equation  x4—  12x2  +  12x=3. 

f  -3.907378, 

,      j  +  .443277, 

AUS'  1  -f  .606018, 

1+2.858083. 

Ex.  4.  Find  the  roots  of  the  equation 

x4-l6x3  +  79x2—  U0x=—  58. 

f  +0.58579, 

j  +3.35425, 

M  +3.41421, 

[  +8.64575. 

Ex.  5.  Find  the  roots  of  the  equation 


+  0.934685, 
+  3.308424, 
Ans  A  +3.824325, 
+4.879508, 
I  +7.053058. 
Ex.  6.  Required  the  fourth  root  of  18339659776. 

Ans.  368. 
Ex.  7.  Required  the  fifth  root  of  26286674882643. 

Ans.  483. 

Ex.  8.  There  is  a  number  consisting  of  four  digits  such  that 
the  sum  of  the  first  and  second  is  9  ;  the  sum  of  the  first  and 
third  is  10  ;  the  sum  of  the  first  and  fourth  is  11  ;  and  if  the 
product  of  the  four  digits  be  increased  by  36  times  the  product 
of  the  first  and  third,  the  sum  will  be  equal  to  3024  diminished 
by  300  times  the  first  digit.  Required  the  number. 

-4  f      6345, 

or  7234, 


or  9012. 


350  ALGEBRA. 

469.  Newton's  Method  of  Approximation. 

Let  x3  +  ~Bxz  +  Cx='V  be  an  equation  to  be  solved. 
by  trial,  a  number,  r,  nearly  equal  to  the  root  sought,  and  let 
r-\-h  denote  the  exact  value  of  the  root,  so  that  h  is  a  small 
fraction  which  is  to  be  determined.  Substitute  r-\-h  for  x  in 
the  given  equation,  and  there  will  result  a  new  equation  con 
taining  only  h  and  known  quantities.  Now,  since  h  is  sup 
posed  to  be  a  small  fraction,  h2  and  h3  will  be  small  compared 
with  h;  and  if  we  reject  the  terms  which  contain  the  second 
and  third  powers  of  /*,  we  shall  have,  approximately, 


-3r2-2Br-CT 

This  correction  applied  to  the  assumed  root  gives  a  closer 
approximation  to  the  value  of  x.  Eepeat  the  operation  with 
this  corrected  value  of  r,  and  a  second  correction  will  be  ob 
tained  which  will  give  a  nearer  value  of  the  root  ;  and,  by  suc 
cessive  repetitions,  the  value  of  the  root  may  be  obtained  to 
any  required  degree  of  accuracy. 

The  value  of  h  may,  however,  be  found  more  briefly  by  ob 
serving  that  the  numerator  is  the  first  member  of  the  equation 
after  V  has  been  transposed  and  x  changed  to  r;  and  the  de 
nominator  is  the  first  derived  function  of  the  numerator  with  a 
negative  sign,  Art.  450. 

EXAMPLES. 

1.  Find  a  root  of  the  equation  x3+2xz+3x=50. 
For  the  numerator  of  the  value  of  A,  we  have 


--  50 

Hence  h=  —    =-=  —  -.  --  —  . 

—8r2—4:r—  3 

We  find,  by  trial,  that  x  is  nearly  equal  to  3.     If  we  substi 
tute  3  for  r,  we  shall  have 


Hence  ce=2.9  nearly.     If  we  substitute  this  new  value  of  r, 
we  shall  find  the  value  of  A  to  be  4-.  00228. 


NUMERICAL   EQUATIONS  OF  HIGHER  DEGREES.  351 

Hence  x=  2.  90228.     If  we  repeat  the  operation  with  this 
last  value  of  r,  we  shall  find  the  value  of  h  to  be  +.0000034 
Hence  #=2.9022834. 

2.  Find  a  root  of  the  equation  x5—  6#=10. 


Here  h= 

Assume  r~2,  and  we  obtain 

A=-^,  or  -0.14. 

Hence  #=1.86  nearly.  If  we  assume  r=1.86,  we  shall  find 
the  value  of  h  to  be  —.021. 

Hence  #=1.839  nearly.  If  we  assume  r—  1.839,  we  shall 
find  the  value  of  A  to  be  +.00001266. 

Therefore  x=  1.83901266. 

3.  Find  a  root  of  the  equation  x3—  9#=  10. 

Ans.  #=3.4494897. 

4.  Find  a  root  of  the  equation  #3  +  9#2+  4^=80. 

Ans.  #=2.4721359. 

470.  Approximation  by  Double  Position.  —  Find,  by  trial,  two 
numbers,  r  and  r',  as  near  as  possible  to  the  true  value  of  x  ; 
substitute  them  successively  for  x  in  the  given  equation,  and 
let  E  and  E'  represent  the  errors  which  result  from  these  sub 
stitutions.  We  assume  that  the  errors  of  the  results  are  pro 
portional  to  the  errors  of  the  assumed  numbers.  This  suppo 
sition  is  not  entirely  correct  ;  but  if  we  employ  numbers  near 
to  the  true  values,  the  error  of  this  supposition  is  generally  not 
very  great,  and  the  error  becomes  less  and  less  the  further  we 
carry  the  approximation.  We  have  then 
E:E'::#-r:#-V. 

Whence,  Art.  305,  E-E'  :  r'—ri:  E  :  x-r; 
that  is,  As  the  difference  of  the  errors  is  to  the  difference  of  the  two 
assumed  numbers,  so  is  either  error  to  the  correction  required  in  the 
corresponding  assumed  number. 

This  correction,  being  added  to  the  assumed  number  when 
it  is  too  small,  or  subtracted  when  too  great,  will  give  a  near 
approximation  to  the  true  root.  This  result,  and  some  other 


352  ALGEBRA. 

number,  may  now  be  used  as  new  values  of  r  and  r'  for  obtain 
ing  a  still  nearer  approximation,  and  so  on. 

It  is  generally  most  convenient  to  assume  two  numbers  which 
differ  only  by  unity  in  the  last  figure  on  the  right,  or  one  of 
the  values  of  r  already  used,  together  with  the  approximate 
root,  may  be  employed  for  the  two  assumed  numbers. 

This  method  of  approximation  is  applicable  to  many  equa 
tions  which  can  not  be  solved  by  either  of  the  preceding 
methods. 

EXAMPLES. 

1.  Find  one  root  of  the  equation  x3+x2+x— 100  =  0. 
When  4  and  5  are  substituted  for  x  in  this  equation,  the  re 
sults  are  —16  and  +55. 

Hence  55  +  16:5—4::  16:. 22. 

Therefore  x^=4.22  nearly. 

We  now  assume  the  two  values  4.2  and  4.3,  and,  substi 
tuting  them  for  x  in  the  given  equation,  we  obtain  the  results 
-4.072  and  +2.297. 

Hence       4.072  +  2.297 :  4.3-4.2  ::  2.297:  .036. 
Therefore  x =4.264  nearly. 

Assuming  again  the  two  values  4.264  and  4.265,  and  sub 
stituting  them  for  x,  we  obtain  the  results  —.027552  and 
+.036535. 

Hence          .064087 :  .001 : :  .027552 :  .0004299. 
Therefore  #=4.2644299  very  nearly. 

2.  Find  one  root  of  the  equation  x3  +  2xz— 23x-70  =  0. 

Ans.  x=5.13458. 

3.  Find  one  root  of  the  equation  x4—3x2— 75^—10000  =  0. 

Ans.  33=10.2610. 

4.  Find  one  root  of  the  equation 

x5  +  3x4  +  2o?3-3x2-2x-2  =  0. 

Ans.  05=1.059109. 

471.  The  different  Roots  of  Unity. — The  equation  xn  —  a  would 
appear  to  have  but  one  root,  that  is,  x  —  \/a;  but,  by  Art.  436, 
it  must  have  n  roots;  that  is,  the  nth  root  of  a  must  have  n  dif 
ferent  values.  Unity  must  therefore  have  two  square  roots, 


NUMERICAL  EQUATIONS  OF  HIGHER  DEGREES.          353 

three  cube  roots,  four  fourth  roots,  five  fifth  roots,  six  sixth 
roots,  and  so  on. 

Ex.  1.  Find  the  two  roots  of  the  equation  x2  —  \. 
Extracting  the  square  root,  we  find  x—  -f-1  or  —1. 
Ex.  2.  Find  the  three  roots  of  the  equation  x3  =  1. 
Since  one  root  of  this  equation  is  x=~L,  the  proposed  equa 
tion  must  be  divisible  by  x—  1  ;  and  dividing,  we  obtain 

cc2+x-f  1  =  0. 
Now  the  roots  of  this  equation  are 

x=— 
Hence  the  required  roots  are 

+  1,  i(-l  +  v^3),  and  i(-l_Z 
which  are  -the  cube  roots  of  unity  ;  and  these  results  may  be 
easily  verified. 

Ex.  3.  Find  the  four  roots  of  the  equation  x*  —  I. 
The  square  root  of  this  equation  is 

x2=-f  1,  or  =-1. 
Hence  the  required  roots  are 

+  1,  -1,  +V^1,  -V^I. 
Ex.  4.  Find  the  five  roots  of  the  equation  cc5  =  l. 
Since  one  root  of  this  equation  is  a?  =  l,  the  proposed  equa 
tion  must  be  divisible  by  x—  1  ;  and  dividing,  we  obtain 


Dividing  again  by  #2,  we  have 


Now  put  v=x-\  —  ,  (2.) 

Ju 


whence  vz—x'2-\-2-\  — 

x2 

which,  being  substituted  in  equation  (1),  gives 

v*  +  v-l  =  Q. 
This  equation,  solved  by  the  usual  method,  gives 


354  ALGEBKA. 

Now  equation  (2)  gives 

a?2— vx=—  1. 


Whence     x=%[v+  Vv2—  4],  and  x=^[v—Vv2—^]9 
from  which,  by  substituting  the  value  of  v,  we  obtain 


and 

Hence  the  fifth  roots  of  unity  are 
1. 


Ex.  5.  Find  the  six  roots  of  the  equation  cc6  =  l. 
These  are  found  by  taking  the  square  roots  of  the  cube  roots 
Hence  we  have 

+  1,  -1,  i±iV^3,  _i±4/^3. 
Ex.  6.  Find  the  four   roots  of  the   equation  x*=—  1,  or 


The  first  member  may  be  made  a  complete  square  by  adding 
2z2;  that  is,  x4  +  2^2  +  1  =  2z2, 

whence  #2  +  1  =  ±  x  \/2  . 

By  transposition  and  completing  the  square, 


Hence 
that  is, 
or 

These  four  values,  together  with  the  four  values  found  in 
Ex.  3,  are  the  eight  roots  of  the  equation 


EXAMPLES   FOR  PRACTICE.  355 


EXAMPLES  FOR  PEACTICE. 

EQUATIONS  OF  THE   FIRST   DEGREE   WITH   ONE   UNKNOWN- 
QUANTITY. 

Ex.1.  Given  12i+3z-6-y=^-5f,  to  find  the  value 
Of  x.  Ans.  x=I39%. 

Ex.2.  Given  a(2x+l$b-10a)=b(x  +  7b),  to  find  tbe  valu8 
of  x.  Ans.  x—6a—7b. 

Ex.  3.  Given  2-^^=1-^^,  to  find  the  value  of  a. 

Ans.  x—4:^. 

/y\  fY* 

Ex.  4.  Given  m-\  —  =n—p—T,  to  find  the  value  of  x. 


a 


T 
o 


(n—p—m)ab 
Ans.  x=±—      -j-2  —  . 
-f  a  +  b 

2^-3    4*-9    8x-27    16x-81      9  , 

Ex.6.  Given  ^g  --  ^0  --  3^"      "2T~  "40'  U 
the  value  of  x.  Ans.  x—Q. 

a4  —  54 

Ex.  6.  Given  a3  +  ^  +  ^2H-^3  =  -    —  ,  to  find  tne  value  of  ^ 

cc 

Ans.  x  —  a—b. 

Ex.7.  Given       z^l_^+l3=26+^,  to  find  the  value 

.  x  —  a—l. 


Ex.  8.  Given  ^-^+^-^=-15,  to  find  the  value  of  x. 
5      10      4       o 

Ans.  x= 


Ex.9.  Given  llix=^+66|-5a?-91,  to  find  the  value 

o 


of  x. 


Ex  10   Given  {5a+^-5  =  -,  to  find  the  value  of  x. 
x  x 

3a-6 

Ans.  x  =  — - — . 
4 


356  ALGEBRA. 


Ex.  11.  Given  c=a+^~  ^,  to  find  the  value  of  x. 

a(m—  3c+3a) 
Ans.  x——^—  —'-. 

c—a-\-m 

Ex.  12.  Given  —  -  ---  -  —  =  —  —  -  to  find  the  value  of  x. 
o  o  44 

Ans.  as=-f. 

Ex.  13.  Given  (m—  x)(n—x)  —  (p-\-  x)(x—q),  to  find  the  val 
ue  of  x. 

mn-4-  pq 

Ans.  x=-  -    -^-^  —  . 
m+n+p—  q 

fio"_99 

Ex.14.  Given  8x-28=(4x-f  21)^      £f,  to  find  the  value 

ox  -f-  14 


Of  #.  . 

Ex.  15.  Given  x=a+-|+^,  to  find  the  value  of  #. 

Ans.  g= 


. 

—cf 

/Y»  /y^y» 

Ex.16.  Given--l-   i+3«J  =  0,  to  find  the  value  of  a:. 


c—  art 

Ex.  17.  Given  (8-3x)2+(4-4x)2^(9-5x)2,  to  find  the  val 
ue  of  x.  Ans.  x=TV 

„  16x+7      05-16      2^  +  1 

Ex.18.  Given  —  —  --  \-  =  —  ^  —  ,  to  find  tbe  value 

£i~t  Lit  -  \)X  O 

of  x.  Ans.  a3=17. 

Ex.19.  Given  ^^-^^=11-1^,  to  find  the  value 

Of  2.  ^.725.05=7. 

-,     OA    r.        l-2x    5-605    8          l-3x2 

Ex.  20.  Gwen      --__—  .  ^  to  find  the 


value  of  x.  -^  Ans.  x—  f. 


EXAMPLES   FOR   PRACTICE.  357 


PROBLEMS  INVOLVING  EQUATIONS  OF  THE  FIRST  DEGREE  WITH 
ONE  UNKNOWN  QUANTITY. 

Prob.  1.  Said  an  old  miser,  For  50  years  I  have  saved  200 
dollars  annually ;  and  for  many  years,  each  of  my  four  sons  has 
saved  annually  the  same  sum,  viz.,  the  oldest  for  27  years  past, 
the  second  since  24  years,  the  third  since  19,  and  the  fourth 
since  16  years.  How  long  since  the  savings  of  the  four  sons 
amounted  in  the  aggregate  to  as  much  as  those  of  the  father? 

Ans.  12  years. 

Prob.  2.  From  four  towns,  A,  B,  C,  D,  lying  along  the  same 
road,  four  persons  start  in  the  stage-coach  for  the  same  place, 
E.  The  distance  from  A  to  B  is  19  miles,  from  B  to  C  3  miles, 
and  from  C  to  D  5  miles.  It  subsequently  appeared  that  the 
person  who  started  from  A  paid  as  much  fare  as  the  three  oth 
er  persons  together ;  and  the  fare  per  mile  was  the  same  for 
each.  It  is  required  to  determine  the  distance  from  D  to  E. 

Ans.  7  miles. 

Prob.  3.  Five  towns,  A,  B,  C,  D,  E,  are  situated  along  the 
same  highway.  The  distance  from  A  to  B  is  37  miles,  from  B 
to  D  34,  and  from  D  to  E  14  miles.  A  merchant  at  C,  situ 
ated  between  A  and  D,  receives  at  one  time  8  tons  of  goods 
from  A,  and  6  tons  from  B.  At  another  time  he  receives  11 
tons  from  D,  and  9  from  E,  and  in  the  latter  case  he  paid  the 
same  amount  for  freight  as  in  the  former,  the  rate  of  transporta 
tion  being  the  same  in  both  cases.  It  is  required  to  compute 
the  distance  from  B  to  C.  "t"~  Ans.  15  miles. 

Prob.  4.  If  20  quarts  of  water  flow  into  a  reservoir  every  3 
minutes,  after  a  certain  time  it  will  still  lack  40  quarts  of  being 
full.  But  if  52  quarts  flow  into  it  every  5  minutes  during  the 
same  period,  72  quarts  of  water  will  have  overflown.  What  is 
the  capacity  of  the  reservoir,  and  how  many  quarts  of  water 
must  flow  into  it  every  minute  in  order  that  it  may  be  just  filled 
in  the  time  before  mentioned? 

Ans.  The  capacity  of  the  reservoir  is  240  quarts,  and 
8  quarts  must  flow  into  it  every  minute. 

Prob.  5.   A  mason,  by  working  10  hours  daily,  could  com- 


358  ALGEBRA. 

plete  in  a  week  as  much  over  888  cubic  feet  of  wall  as  at  pres 
ent  he  completes  less  than  888  cubic  feet,  working  only  8-J 
hours  daily.  How  many  cubic  feet  of  wall  does  he  now  com 
plete  weekly  ?  Ans.  816  cubic  feet. 

Prob.  6.  After  a  certain  time  I  have  $670  to  pay,  and  4J 
months  later  I  have  $980  to  pay.  I  settle  both  bills  at  once, 
at  4f  per  cent,  discount,  for  $1594.41.  When  did  the  first  sum 
become  due?  Ans.  After  5£  months. 

Prob.  7.  A  merchant  gains  8  per  cent,  when  he  sells  a  hogs 
head  of  oil  at  36  dollars.  How  much  per  cent,  does  he  gain  or 
lose  when  he  sells  a  hogshead  at  32  dollars? 

Ans.  He  loses  4  per  cent. 

Prob.  8.  A  merchant  loses  2J  per  cent,  when  he  sells  a  bag 
of  coffee  for  39  dollars.  How  much  per  cent,  does  he  gain  or 
lose  when  he  sells  a  bag  of  coffee  for  41 J-  dollars  ? 

Ans.  He  gains  3f  per  cent. 

Prob.  9.  A  merchant  owes  $2007,  to  be  paid  after  5  months, 
$3395  after  7  months,  and  $6740  after  13  months.  When 
should  the  entire  sum  of  $12,142  be  paid,  so  that  neither  party 
may  sustain  any  loss?  Ans.  After  10  months. 

Prob.  10.  A  merchant  has  three  sums  of  money  to  pay,  viz., 
$1013  after  3J  months,  $431  four  months  later,  and  the  third 
sum  still  four  months  later.  How  large  is  the  third  sum,  sup 
posing  he  could  pay  the  three  bills  together  in  6J  months  with 
out  loss  or  gain  ?  V-  Ans.  $428. 

Prob.  11.  A  merchant  has  two  kinds  of  tobacco ;  the  one  cost 
40  cents  per  pound,  the  other  24  cents.  He  wishes  to  mix  the 
two  kinds  together,  so  that  he  may  sell  it  at  34  cents  per  pound 
without  loss  or  gain.  How  much  must  he  take  of  each  sort  in 
order  to  have  64  pounds  of  the  mixture? 

Ans.  40  pounds  of  the  better  sort,  and  24  pounds  of 
the  poorer. 

Prob.  12.  A  vinegar  dealer  wishes  to  dilute  his  vinegar  with 
water.  At  present  he  sells  his  vinegar  at  6  dollars  per  hogs 
head  (120  quarts).  How  much  water  must  he  add  to  29^ 
hogsheads  in  order  to  be  able  to  sell  the  mixture  at  4  cents  per 
quart?  Ans.  7f  hogsheads. 


EXAMPLES   FOR   PRACTICE.  359 

Prob.  13.  A  metallic  compound  consists  of  4  parts  copper 
and  3  parts  silver.  How  much  copper  must  be  added  to  94J 
pounds  of  the  compound,  in  order  that  the  proportions  may  be 
7  parts  of  copper  to  2  parts  of  silver  ?±  Ans.  87f  pounds. 

Prob.  14.  In  255  pounds  of  spirit  of  wine,  water  and  pure  al 
cohol  are  combined  by  weight  in  the  ratio  of  2  to  3.  How 
much  water  must  be  extracted  by  distillation,  in  order  that  the 
ratio  of  the  water  to  the  alcohol  may  be  3  to  17  by  weight? 

Ans.  75  pounds. 

Prob.  15.  It  is  required  to  diminish  each  of  the  factors  of  the 
two  unequal  products,  52x45  and  66x37,  by  the  same  num 
ber,  so  that  the  new  products  may  be  equal  to  each  other. 
What  is  that  number?  Ans.  17. 

Prob.  16.  The  square  of  a  certain  number  is  1188  greater  than 
the  square  of  a  number  smaller  by  6  than  the  former.  What 
is  that  number?  Ans.  102. 

Prob.  17.  I  have  a  certain  number  of  dollars  in  my  posses 
sion,  which  I  undertook  to  arrange  in  the  form  of  a  square,  and 
found  that  I  wanted  25  dollars  to  complete  the  square ;  but  if 
I  diminish  each  side  of  the  square  by  2,  there  remain  31  dollars 
over.  How  many  dollars  have  I?  Ans.  200. 

Prob.  18.  A  vine-tiller  has  a  rectangular  garden,  whose 
length  is  to  its  breadth  as  7  to  5,  which  he  wishes  to  plant  with 
vines.  If  he  sets  the  plants  at  a  certain  uniform  distance  from 
each  other,  he  finds  that  he  has  2832  plants  remaining.  But 
if  he  places  them  nearer  together,  so  as  to  make  14  more  on 
each  longer  side,  and  10  more  on  each  shorter  side,  he  has  only 
172  plants  remaining.  How  many  plants  has  he? 

^  Ans.  14,172. 

Prob.  19.  In  the  composition  of  a  certain  quantity  of  gun 
powder,  the  nitre  was  ten  pounds  more  than  two  thirds  of  tho 
whole ;  the  sulphur  was  four  and  a  half  pounds  less  than  one 
sixth  of  the  whole ;  and  the  charcoal  was  two  pounds  less  than 
one  seventh  of  the  nitre.  How  many  pounds  of  gunpowder 
were  there  ?  Ans.  69  pounds. 

Prob.  20.  There  are  three  numbers  in  the  ratio  of  3,  4,  and 
5.  Five  times  the  first  number,  together  with  four  times  the 


362 


ALGEBRA. 


Ex.  11. 


-.     . 

~~ '  T— 

a     o 

X       Z 

-+-=4 
a     c 


(5x— 6^  +  42-15 
Ex.12.   ^7x  +  4y-32-19 


fcc 


-21— 


Ex.13.  ^  *  =  9-y 


Ans.       =2&. 


—  4. 


Ex.14,  1  Sx—  5?/  +  72—  75V 
(  9x  —  112  +  10  —  0    ) 

Ans.  \y=—5. 
(2=6. 

(  3x  —  5?/  +  42  —  5  } 
Ex.15,   -j  lx  +  2y—  32  —  2  t 

^J«-2' 

(4x  +  3?/—  2  —  7  ) 

(z=S. 

Ex.16.  •<  2x  +  3^—42—20  V 

An,.\y=i 

v  ox  —  ^y  +  O2^^^o  / 

(  2  —  2. 

f       t-r               o               ~f    ^ 

f  7^-%=n 

Ex.l7.-|  £l-ll-lt 

j  y—^- 

Ans'\  2-16. 

[t*=25. 

J~ 

fx-12 

T?     10    j  2x  +  3y—  39  f 

iilX.      lO.          -<         ,;                         f-                       -.-if 

5x  —  72—  11 

Ans-\  Si 

[4?/+32—  41  J 

[w=4. 

T 

Ex.19.  j!;^2x-30         f 

1                                                             | 

^|s 

1  w-9. 

EXAMPLES   FOR  PRACTICE.  363 


=105]  fa?=30. 

20  =317  An* 

'20'  u  -741  f  ^ 

J 

-     2. 


Ex.21,  -j  lOy—  3a  +  3w-2v  =  2  Ans.     z=     3. 

2y-2x=3  u=—l. 


PROBLEMS  INVOLVING  EQUATIONS  OF  THE  FIRST  DEGREE  WITH 
SEVERAL  UNKNOWN  QUANTITIES. 

Prob.  1.  Two  sums  of  money,  which  were  put  out  at  inter 
est,  the  one  at  5  per  cent,  the  other  at  4J  per  cent.,  yielded  in 
one  year  $284.40  interest.  If  the  former  sum  had  been  put  out 
at  4f  per  cent.,  and  the  latter  at  5  per  cent.,  they  would  have 
yielded  $4.50  less  interest.  What  were  the  two  sums  of  money? 
Ans.  One  was  $3420,  the  other  $2520. 

Prob.  2.  There  is  a  number  consisting  of  two  digits;  the 
number  is  equal  to  three  times  the  sum  of  its  digits,  and  if  it  be 
multiplied  by  three,  the  result  will  be  equal  to  the  square  of  the 
sum  of  its  digits.  Find  the  number.  ^  Ans.  27. 

Prob.  3.  A  merchant  sold  two  bales  of  goods  for  the  sum  of 
$987-|,  the  first  at  a  loss  of  8f  per  cent,  the  second  at  a  loss  of 
11J  per  cent  If  he  had  sold  the  first  at  a  loss  of  11  J  per  cent, 
and  the  second  at  a  loss  of  8f-  per  cent.,  he  would  have  received 
the  sum  of  $992f.  How  much  did  each  bale  cost? 

Ans.  The  first  $455,  the  second  $645. 

Prob.  4.  Two  messengers,  A  and  B,  from  two  towns  distant 
57-J  miles  from  each  other,  set  out  to  meet  each  other.  If  A 
starts  5f  hours  earlier  than  B,  they  will  meet  in  6J-  hours  after 
B  starts;  but  if  B  starts  5f  hours  earlier  than  A,  they  will  meet 
in  5f  hours  after  A  starts.  How  many  miles  does  each  travel 
in  an  hour?  -K  Ans.  A  3  miles,  and  B  3^  miles. 

Prob.  5.  A  jeweler  has  two  masses  of  gold  of  different  de 
grees  of  fineness.  If  he  mixes  10  ounces  of  the  one  with  5 
ounces  of  the  other,  he  obtains  gold  11  carats  fine;  but  if  he 


364  ALGEBRA. 

mixes  7-J-  ounces  of  the  former  with  1J-  ounces  of  the  latter,  he 
obtains  a  mixture  10  carats  fine.  What  was  the  fineness  of 
each  mass  ?  Ans.  The  one  9  carats,  the  other  15  carats. 

Prob.  6.  A  farmer  has  a  certain  number  of  oxen,  and  proven 
der  for  a  certain  number  of  days.  If  he  sells  75  oxen,  his  prov 
ender  will  last  20  days  longer ;  but  if  he  buys  100  more  oxen, 
his  provender  will  be  exhausted  15  days  sooner.  How  many 
oxen  has  he,  and  how  many  days  will  the  provender  last  ? 

Ans.  300  oxen,  and  the  provender  will  last  60  days. 

Prob.  7.  A  certain  number  of  laborers  remove  a  pile  of  stones 
in  6  hours  from  one  place  to  another.  If  there  had  been  2  more 
laborers,  and  if  each  laborer  had  each  time  carried  4  pounds 
more,  the  pile  would  have  been  removed  in  5  hours ;  but  if  there 
had  been  3  less  laborers,  and  if  each  laborer  had  each  time  car 
ried  5  pounds  less,  it  would  have  required  8  hours  to  remove 
the  pile.  How  many  laborers  were  there,  and  how  much  did 
each  carry  at  one  time? 

Ans.  There  were  18  laborers,  and  each  carried  50  pounds. 

Prob.  8.  A  heavy  wagon  requires  a  certain  time  to  travel 
from  A  to  B.     A  second  wagon,  which  every  4  hours  travels  5 
miles  less  than  the  first,  requires  4  hours  more  than  the  first  to 
go  from  A  to  B.     A  third  wagon,  which  every  3  hours  travels 
8f  miles  more  than  the  second,  requires  7  hours  less  than  the 
second  to  make  the  same  journey.     How  far  is  A  from  B,  and 
what  time  does  each  wagon  require  to  travel  this  distance? 
Ans.  From  A  to  B   is   60   miles;    the  first  wagon 
requires  12  hours,  the  second  16,  and  the  third 
9  hours. 

Prob.  9.  I  have  two  equal  sums  to  pay,  one  after  9,  and  the 
other  after  15  months.  If  I  settle  them  both  at  once,  at  the 
same  rate  of  discount,  I  must  pay  for  the  first  sum  $1208,  and 
for  the  second  $1160.  How  much  was  each  sum,  and  at  what 
per  cent  was  the  discount  reckoned  ? 

Ans.  $1280,  and  the  discount  was  7-J  per  cent. 

Prob.  10.  A  small  square  lies  with  one  angle  in  the  angle  of  a 
larger  square.  The  excess  of  the  side  of  the  larger  square  above 
that  of  the  smaller  is  118  feet;  the  excess  of  the  square  itself 


EXAMPLES  FOR  PRACTICE.  365 

is  26,432  square  feet.  What  are  the  contents  of  each  of  the 
two  squares  ? 

Ans.  The  one  29,241,  the  other  2809  square  feet. 
Prob.  11.  It  is  required  to  find  two  numbers  whose  sum,  dif 
ference,  and  product  are  in  the  ratio  of  the  numbers  5, 1,  and  18. 

Ans.  9  and  6. 

Prob.  12.  Two  numbers  are  in  the  ratio  of  7  to  3,  and  their 
difference  is  to  their  product  as  1  to  21.  What  are  the  num 
bers?  -dm  28  and  12. 

Prob.  13.  Three  towns,  A,  B,  and  C,  lie  at  the  angles  of  a 
triangle.  From  A  by  B  to  C  is  164  miles ;  from  B  bj  C  to  A 
is  194  miles ;  and  from  C  by  A  to  B  is  178  miles.  How  far 
are  A,  B,  and  C  from  each  other  ? 

Ans.  From  A  to  B  74  miles,  from  B  to  C  90,  and  from 

C  to  A  104  miles. 

Prob.  14.  A  railway  train,  after  traveling  for  one  hour,  meets 
with  an  accident  which  delays  it  one  hour,  after  which  it  pro 
ceeds  at  three  fifths  of  its  former  rate,  and  arrives  at  the  termi 
nus  three  hours  behind  time ;  had  the  accident  occurred  50 
miles  further  on,  the  train  would  have  arrived  1  hour  and  20 
minutes  sooner.  Required  the  length  of  the  line. 

Ans.  100  miles ;  original  rate  25  miles  per  hour. 
Prob.  15.  A  railway  train,  running  from  New  York  to  Al 
bany,  meets  on  the  way  with  an  accident,  which  causes  it  to 

diminish  its  speed  to  -th  of  what  it  was  before,  and  it  is  in  con 
sequence  a  hours  late.  If  the  accident  had  happened  b  miles 
nearer  Albany,  the  train  would  have  been  c  hours  late.  Find 
the  rate  of  the  train  before  the  accident  occurred. 

b(n—I)      ., 

Ans.  — '-  miles  per  hour. 

a—c 

Prob.  16.  Three  boys  are  playing  with  marbles.  Said  A  to 
B,  Give  me  5  marbles,  and  I  shall  have  twice  as  many  as  you 
will  have  left.  Said  B  to  C,  Give  me  13  marbles,  and  I  shall 
have  three  times  as  many  as  you  will  have  left.  Said  C  to  A, 
Give  me  3  marbles,  and  I  shall  have  six  times  as  many  as  you 
will  have  left.  How  many  marbles  had  each  boy  ? 

Ans.  A  had  7,  B  11,  and  C  21  marbles. 


366  ALGEBRA. 

Prob.  17.  It  is  required  to  divide  the  number  232  into  three 
parts  such  that,  if  to  the  first  we  add  half  the  sum  of  the  oth 
er  two,  to  the  second  we  add  one  third  the  sum  of  the  other 
two,  and  to  the  third  we  add  one  fourth  the  sum  of  the  other 
two,  the  three  results  thus  obtained  shall  be  equal.  What  are 
the  parts  ? 

Ans.  The  first  40,  the  second  88,  and  the  third  104 

Prob.  18.  Four  towns,  A,  B,  C,  and  D,  are  situated  at  the 
angles  of  a  quadrilateral  figure.  When  I  travel  from  A  by  B 
and  C  to  D,  I  pay  $6.10  passage-money;  when  I  travel  from 
A  by  D  and  C  to  B,  I  pay  $5.50.  From  A  by  B  to  C,  I  pay 
the  same  as  from  A  by  D  to  C ;  but  from  B  by  A  to  D,  I  pay 
40  cents  less  than  from  B  by  C  to  D.  What  are  the  distances 
of  the  four  towns  from  each  other,  supposing  I  paid  in  each  case 
10  cents  per  mile  ? 

Ans.  From  A  to  B  21,  from  B  to  C  17,  from  C  to  D 
23,  and  from  D  to  A  15  miles. 

Prob.  19.  Four  players,  A,  B,  C,  and  D,  play  four  games  at 
cards.  At  the  first  game  A,  B,  and  C  win,  and  each  of  them 
doubles  his  money ;  at  the  second  game  A,  B,  and  D  win,  each 
of  them  doubling  the  money  he  had  at  the  commencement  of 
that  game;  at  the  third  game  A,  C,  and  D  win;  and  at  the 
fourth  game  B,  C,  and  D  win  ;  and  at  each  game  each  winner 
won  as  much  money  as  he  had  at  the  commencement  of  that 
game.  They  now  count  their  money,  and  find  that  each  has 
$64.  How  much  had  each  before  commencing  play? 
'/  Ans.  A  had  $20,  B  had  $36,  C  had  $68,  and  D  had  $132. 

Prob.  20.  A  and  B  start  together  from  the  foot  of  a  mountain 
to  go  to  the  summit.  A  would  reach  the  summit  half  an  hour 
before  B,  but,  missing  his  way,  goes  a  mile  and  back  again  need 
lessly,  during  which  he  walks  at  twice  his  former  pace,  and 
jreaches  the  top  six  minutes  before  B.  C  starts  twenty  minutea 
after  A  and  B,  and,  walking  at  the  rate  of  two  and  one  seventh 
miles  per  hour,  arrives  at  the  summit  ten  minutes  after  B. 
Find  the  rates  of  walking  of  A  and  B,  and  the  distance  from 
the  foot  to  the  summit  of  the  mountain. 

Ans.  2^,  2 ;  distance  5  miles. 


EXAMPLES   FOR   PRACTICE.  367 

Prob.  21.  Find  three  numbers  such  that  if  six  be  subtracted 
from  the  first  and  second,  the  remainders  will  be  in  the  ratio 
of  2  :  3  ;  if  thirty  be  added  to  the  first  and  third,  the  sums  will 
be  in  the  ratio  of  3 : 4 ;  but  if  ten  be  subtracted  from  the  sec 
ond  and  third,  the  remainders  will  be  as  4 :  5. 

-f-  Ans.  30,  42,  50. 

Prob.  22.  A  and  B  engage  to  reap  a  field  of  wheat  in  twelve 
days.  The  times  in  which  they  could  severally  reap  an  acre 
are  as  2:3.  After  some  days,  finding  themselves  unable  to 
finish  it  in  the  stipulated  time,  they  call  in  C  to  help  them, 
whose  rate  of  working  was  such  that,  if  he  had  wrought  with 
them  from  the  beginning,  it  would  have  been  finished  in  nine 
days.  Also,  the  times  in  which  he  could  have  reaped  the  field 
with  A  alone,  and  with  B  alone,  are  in  the  ratio  of  7 :  8.  When 
was  C  called  in?  Ans.  After  six  days. 

EQUATIONS  OF  THE   SECOND   DEGREE  WITH  ONE   UNKNOWN 

QUANTITY. 

A. — INCOMPLETE    EQUATIONS    OP   THE    SECOND   DEGREE. 

-r,       _,      ri.  X  +  I8      X— 18      5    ,  f 

Ex.  1.  Given —  H —= -,  to  find  the  values  of  x. 

T>_u9         T      9       V 

i//  "l™  £i  & "™~-  £i  O 

Ans.  x=  ±14. 

Ex.2.  Given \/4+49—\/4~49  =  7,  to   find   the   values 
v  x2  v  x2 

of  x.  Ans.  x—  ±f. 

Ex.3.  Given  -  +  -=-  +  -  to  find  the  values  of  a?. 
•     5     x     2     x 

Ans.  x=db  VlO. 
Ex.  4.  Given  x-\-  -\/a + x2 — ;  to  find  the  values  of  x. 


Ans.  x  =  ±  -J  (a  —  1). 


Ex.5.  Giveny^  +  m2-3  =  m  +  l-y^—  2,  to  find  the 
values  of  x.  Ans.  x=  ±m. 

Ex.6.  Given  y^+29-\/^-34=7,  to  find  the  val 
ues  of  x.  Ans.  x  = 


868 


Ex.  7.  Given 
ues  of  x. 


ALGEBRA. 


=—  IT,  to  find  the  val 


fVl-X2 

Ex.8.  Given  27(7-x)2-43  =  77-3(7-x)2,  to  find  the  val 
ues  of  x. 

Remark.  Put  7—x=y;  first  find  the  value  of  #,  and  thence  the  value  of  x. 

Ans.  a?=5  or  9. 
Ex.  9.  Given  -       ,    ~"  =5,  to  find  the  values  of  x. 


a+Va2—  x 


.Ans.  x  =  ± 


Ex.  10.  Given 


Vx—Vx—a    x—a 


find 


Ans'  -ra 


Ex.11.  Given 


Ex.12.  Given 


V 


X 


V.27 


=\y,  to  find  the  values  of  07. 


+X 


,  to  find  the  values  of  x. 


Vl  +  x     1  —  VI—  x 

j\  Ans.  x= 

B.  —  COMPLETE  EQUATIONS  OF  THE  SECOND  DEGREE. 

Ex.  13.  Given  557x=5801i+8x2,  to  find  the  values  of  x. 

Ans.  x=56£  or  12  J. 
Ex.  14.  Given  (7x)2—  7x=l,  to  find  the  values  of  x. 

Ans.  x=0.2311477  or  —0.0882905. 

Ex.  15.  Given  12x2=2H-  Jx,  to  find  the  values  of  x. 

Ans.  x  =  l^  or  —  1TV 

Ex.16.  Given  57x—18x2+  145  =  0,  to  find  the  values  of  x. 

Ans.  x=4£  or  —If. 

Ex.17.  Given   ^(x+l)-l(2x*+x-I)  =  ^-(x+l),  to  find 
^51  /  oO 

the  values  of  x.  Ans.  x—  —  1  or  f  . 


EXAMPLES   FOK  PRACTICE.  369 


Ex.  18.  Given  —  ^r=s,  to  find  the  values  of  x- 

x—I2      x—Q      6 

Ans.  x—24:  or  ^-. 

X-4-4:       X 4       10 

Ex.  19.  Given --\ r  =  -s~,  to  find  the  values  of  x. 

X  —  4:       X  +  4         3 

Ans.  x=8  or  —8. 
id  the  values  of  x. 
Ans.  x=o  or  — |-. 

£      *W  1  ft      T 

TT1         ^^^        /^  •  a  ~T~  ^^^          "  —  c'c*y  -L^-'  —  •**  .          r*      j      AT_  i 

Ex.  21.  Given  _3__^_lg  =  __lsg,  to  find  the  val- 
ues  of  x.  Ans.  0^=8  or  — 2-j-y^-- 

Ex.  22.  Given  -   — --f  -      — =-,  to  find  the  values  of  x. 
ox — o     Zx — o     A        , 

Ans.  x=.^  or  1. 

Ex.  23.  Given --\ T=— - — ~r.  to  find  the  values  of  x. 

x—I     x—l       x—3 


Ex.  20.  Given  -  -  H  --  -=  -  -,  to  find  the  values  of  x. 
x+l     x+2     ce+3' 


^?i5.  x=     or  — 

Ex.24.  Given  (7-4V3)^2+(2-  V~3)x=2,  to  find  the  val 
ues  of  x.  Ans.  x=2+V$,  or  —2(2  +  -v/3). 

^/r  91  _  1/^r 

Ex.25.  Given  —  ^—  +        x-     =2|,  to  find  the  values  of  x. 
21  —  Vx         Vx 

Ans.  #=:49  or  196. 

Ex.  26.  Given  \/x+  Vx=20,  to  find  the  values  of  x. 
Ans.  cc=+44=256  or   -54  = 


Ex.27.  Given  <  -  -  -  =_+_+-  to  find  the  values  ofx. 
a-fo-j-cc     a     o     x 

Ans.  x=—a  or  —b. 

Ex.  28.  Given          -^^J^±  to  find  the  values  of  x. 
a+x     a—b 

Ans.  g== 


. 

2 

Ex.  29.  Given  =         find 

a  +  x 

Q2 


370  ALGEBRA. 


Ex.30.  Given  x*—  4x3  +  7x2  —  60;  =  18,  to  find  the  values  of  9 
by  a  quadratic  equation.  Ans.  x  =  3  or  —  1,  or  1±  V—  5. 

PROBLEMS   INVOLVING   EQUATIONS   OF   THE   SECOND   DEGREE 
WITH   ONE    UNKNOWN   QUANTITY. 

Prob.  1.  It  is  required  to  find  three  numbers  which  are  in  the 
ratio  of  J,  -J,  and  J,  and  the  sum  of  whose  squares  is  10,309. 

Ans.  78,  52,  39. 

Prob.  2.  A  gentleman  buys  a  certain  number  of  pounds  of 
salt,  four  times  as  much  sugar,  and  eight  times  as  much  coffee, 
and  for  each  pound  of  the  three  articles  he  paid  as  many  cents 
as  he  bought  pounds  of  that  article.  For  the  whole  he  paid 
$3.24.  How  many  pounds  of  coffee  did  he  buy? 

Aits.  16  pounds. 

Prob.  3.  A  rectangular  garden  was  37  feet  broad  and  259  feet 
long.  Its  breadth  was  increased  by  a  certain  number  of  feet, 
and  its  length  diminished  by  seven  times  that  number,  by  which 
means  its  area  was  diminished  63  square  feet.  By  how  many 
feet  was  the  breadth  increased  ?  Ans.  3  feet. 

Prob.  4.  Find  that  number  whose  square  added  to  its  cube 
is  nine  times  the  next  higher  number.  Ans.  3. 

Prob.  5.  A  sets  out  from  New  York  to  Chicago,  and  B  at  the 
same  time  from  Chicago  to  New  York,  and  they  travel  uniform 
ly  ;  A  reaches  Chicago  16  hours,  and  B  reaches  New  York  36 
hours  after  they  have  met  on  the  road.  Find  in  what  time 
each  has  performed  the  journey. 

Ans.  A  40  hours,  B  60  hours. 

Prob.  6.  A  square  vineyard,  in  which  the  vines  are  set  in 
squares  so  as  to  be  uniformly  four  feet  apart,  is  to  be  replanted 
so  that  the  vines  may  be  uniformly  3J  feet  apart.  Supposing 
8640  more  vines  are  required  for  this  change,  what  must  be  the 
length  of  each  side  of  the  vineyard?  Ans.  672  feet. 

Prob.  7.  A  glass  mirror,  33  inches  high  and  22  inches  wide, 
is  to  be  set  in  a  frame  of  uniform  breadth,  such  that  the  surface 
of  the  frame  shall  be  just  equal  to  that  of  the  glass.  What 
must  be  the  breadth  of  the  frame?  Ans.  5J-  inches. 

Prob.  8.  Required  the  solution  of  the  preceding  problem,  if 


EXAMPLES   FOR   PRACTICE.  371 

we  represent  the  height  of  the  mirror  by  a  and  its  breadth  by  6, 
and  it  is  required  that  the  surface  of  the  frame  shall  be  p  times 
that  of  the  mirror. 


• 
4: 

Prob.  9.  Sixty  pounds  of  a  certain  quality  of  sugar  cost  $2.40 
less  than  sixty  pounds  of  another  quality.  If  I  buy  sugar  of 
each  quality  to  the  amount  of  $5.04,  1  obtain  of  the  first  kind 
8  pounds  more  than  of  the  second.  What  was  the  price  of  a 
pound  of  each  kind  ? 

Ans.  One  14  cents,  the  other  18  cents. 

Prob.  10.  A  gentleman  bought  a  horse  for  a  certain  sum. 
He  afterward  sold  him  for  $144,  and  gained  as  much  per  cent, 
as  the  horse  cost  him.  How  much  did  he  pay  for  the  horse  ? 

Ans.  80  dollars. 

Prob.  11.  A  merchant  buys  a  certain  number  of  barrels  of 
flour  for  $216.  At  another  time  he  expended  the  same  sum  of 
money  for  flour,  but  obtained  three  barrels  less,  the  price  of 
flour  having  risen  one  dollar  per  barrel.  How  many  barrels 
did  he  buy  in  the  first  case?  Ans.  27  barrels. 

Prob.  12.  A  and  B  contribute  together  $3400  in  trade,  A  for 
12  and  B  for  16  months.  In  the  distribution,  A  received  $2070, 
capital  and  profits,  and  B  received  $1920.  What  was  each  one's 
capital?  Ans.  A  contributed  $1800,  and  B  $1600. 

Prob.  13.  Supposing  the  mass  of  the  earth  to  be  80  times  that 
of  the  moon,  their  distance  240,000  miles,  and  the  force  of  at 
traction  to  vary  directly  as  the  quantity  of  matter,  and  inverse 
ly  as  the  square  of  the  distance,  at  what  point  between  them 
will  a  third  body  be  equally  attracted  by  the  earth  and  moon  ? 

Ans.  24,134  miles  from  the  moon. 

Prob.  14.  A  wall  was  completed  in  5-J  days  by  two  masons, 
one  of  whom  commenced  work  1-J-  days  later  than  the  other. 
In  order  to  complete  the  wall  alone,  the  first  would  have  re 
quired  3  days  less  than  the  second.  In  how  many  days  could 
each  alone  complete  the  wall  ? 

Ans.  The  first  in  8,  the  second  in  11  days. 

Prob.  15.  A  courier  goes  from  a  place,  A,  to  a  place,  B,  io 


372  ALGEBRA. 

14  hours.  At  the  same  time,  another  courier  starts  for  B  from 
a  place  10  miles  further  distant,  and  expects  to  reach  B  at  the 
same  time  with  the  first,  by  gaining  half  an  hour  in  every  20 
miles.  What  is  the  distance  from  A  to  B  ?  Ans.  70  miles. 

Prob.  16.  From  two  towns,  A  and  B,  which  are  104  miles 
distant  from  each  other,  two  wagons  start  at  the  same  time, 
and  meet  after  10-J  hours.  One  requires  for  every  8  miles  a 
quarter  of  an  hour  more  than  the  other.  How  much  time  does 
each  require  to  travel  one  mile  ? 

Ans.  The  one  -£%,  the  other  -^  of  an  hour. 

Prob.  17.  Two  messengers  start  at  the  same  time  from  two 
towns,  A  and  B,  the  first  toward  B,  the  other  toward  A,  and, 
upon  meeting,  it  appeared  that  the  first  had  traveled  12  miles 
more  than  the  second ;  also,  that  if  each  should  continue  on  at 
his  former  rate,  the  first  would  arrive  at  B  in  9  hours,  and  the 
latter  at  A  in  16  hours.  What  is  the  distance  from  A  to  B? 

Ans.  84  miles. 

Prob.  18.  Two  messengers  start  from  the  two  towns,  A  and 
B,  to  travel  toward  each  other,  but  one  started  two  hours  ear 
lier  than  the  other.  They  meet  each  other  2^  hours  after  the 
starting  of  the  second  messenger,  and  they  reach  the  towns  A 
and  B  at  the  same  instant.  In  how  many  hours  did  each  mes 
senger  perform  the  journey  ? 

Ans.  The  one  in  7,  the  other  in  5  hours. 

Prob.  19.  Two  travelers  start  from  two  towns,  A  and  B, 
whose  distance  from  each  other  is  910  miles,  and  travel  uni 
formly  toward  each  other.  If  the  first  starts  56  hours  before 
the  second,  they  will  meet  halfway  between  A  and  B.  If  both 
start  at  the  same  instant,  at  the  end  of  20  hours  they  will  still 
be  550  miles  from  each  other.  How  many  hours  does  each 
traveler  require  to  accomplish  the  distance  from  A  to  B  ? 

Ans.  One  182  hours,  the  other  70  hours. 

Prob.  20.  A  grocer  has  a  cask  containing  20  gallons  of  bran 
dy,  from  which  he  draws  off  a  certain  quantity  into  another 
cask  of  equal  size,  and,  having  filled  the  last  with  water,  the 
first  cask  was  filled  with  the  mixture.  It  now  appears  that  if 
6|  gallons  of  the  mixture  are  drawn  off  from  the  first  into  the 


EXAMPLES   FOR   PRACTICE.  373 

second  cask,  there  will  be  equal  quantities  of  brandy  in  each, 
Eequired  the  quantity  of  brandy  first  drawn  off. 

Ans.  10  gallons. 

Prob.  21.  Two  merchants  sold  the  same  kind  of  cloth.  The 
second  sold  three  yards  more  of  it  than  the  first,  and  together 
they  received  $35.  The  first  said  to  the  second,  I  should  have 
received  $24  for  your  cloth ;  the  other  replied,  I  should  have 
received  $12 j-  for  yours.  How  many  yards  did  each  of  them 
sell? 

Ans.  The  first  merchant  5  or  15  yards,  the  second 
merchant  8  or  18  yards. 

Prob.  22.  A  and  B  traveled  on  the  same  road,  and  at  the 
same  rate,  from  Cumberland  to  Baltimore.  At  the  50th  mile 
stone  from  Baltimore  A  overtook  a  drove  of  geese,  which  were 
proceeding  at  the  rate  of  three  miles  in  two  hours,  and  two 
hours  afterward  met  a  wagon,  which  was  moving  at  the  rate 
of  nine  miles  in  four  hours.  B  overtook  the  same  drove  of 
geese  at  the  45th  milestone,  and  met  the  same  wagon  40  min 
utes  before  he  came  to  the  31st  milestone.  Where  was  B  when 
A  reached  Baltimore?  Ans.  25  miles  from  Baltimore. 

w 

EQUATIONS  OF  THE  SECOND  DEGREE  WITH  SEVERAL  UNKNOWN 

QUANTITIES. 

Ex.1.  Given  (13x)2  +  2?/2  =  177, )  to   find  the   values   of  x 
(2y)2-I3x*  =     3,  |  and  y. 

Ans.  ce=  ±1,  y  —  ±2. 

Ex.2.  Given  x^  +  fix2— f-.\  25:7, )  to  find  the  values  of 
xy  —  4:S,  f  x  and  y. 

Ans.  x=  ±8,  y=±6. 

Ex.  3.  Given  2(cc+4)2—  5(y— 7)*=     75, )  to  find  the  values 
7(>+4)2  +  15(?/-7)2:=1075,  j-       ofx  and  ^ 

Remark.  Put  x  +  ±=z,  and  y  —  7—v.  First  determine  z  and  v,  and  thence  x 
and  y. 

Ans.  x=  +6  or  —14,  y  =  12  or  2. 

Ex.4.  Given  (x  +  y)*  —  2x2  -  49,)  to  find  the  values  of  x 

=  372,  j  and  y. 

Ans.  x—  ±4,  y=  ±5,  or  ±13. 


374  ALGEBRA. 

Ex.5.  Given  2cc-l-3?/  =  37J 


11      _14  >  to  find  the  values  of  x  and 
ay     ~45'  ) 

.  x  =5  or  -3-     =  9  or  -7 


Ex.6.  Given  -  +     =9, 

y      x          V  to  find  the  values  of  x  and  y. 


.  x=4:  or  2,  y=2  or  4. 

Ex.  7.  Given  x2  +  7/2  =  10000, 

' 


vJ  J-  to  find  the  values  of  x  and  ?/ 
y  -     124,  ) 

^ns.  x=96  or  28,  y  =  28  or  96. 

Ex.  8.  Given  12  :  x  : :  y  :  3, )  .  f 

*        I  to  find  the  values  of  x  and  y. 
yx-\-  vy  =  5,  ) 

Ans.  x=9  or  4,  y=4:  or  9. 

Ex.9.  Given  (3x+4?/)(7x— 2?/)  +  3x+4?/=.44,  )  to   find   the 
(3x+4?/)(7x — 2y) — 7cc  +  2?/  =  30,  f  values  of  x 


and  y.  J.?w.  x  =  l  or  1T7T,  y  =  2  or  — iV- 

Ex.10.  Given  —  x2+6xz/-9?/24-4x— 12y=  4,)  to  find  the 
x2— 2xy  f  By2— 4x  +  5y  =53,  f  values  of  x 
and  y.  Ans.  x  —  ll  or  —  7-J-,  y  =  3  or  —  3J. 

Ex.  11.  Given  2(x2  +  3/2)(x-f?/)    =15a#,     )  to  find  the  values 
4(x4 — y4)(x2 — 2/2)=45x2y3,  j        ofccand?/. 
^dw5.  x— 2  or  1,  y—~L  or  2. 

Ex.12.  Given  (x2  —  y*)(x—y)    =   16a?y,     )  to  find  the  values 


Ans.  x=9  or  3,  ^/^=3  or  9. 
Ex.13.  Given  jc(a5+2M-z)  =  27, 


Ex.14.  Given  xy=z,  ~\ 

/        -  st 


to  find  the  vf  ues  of  x, 
and,. 

Ans.  x  =  3,  ?/  =  2,  2=4. 


}-  to  find  the  values  of  x,  y,  z,  and  v. 
—  ct, 


yv  =  o 

Va         3/-  /-  /-3/- 


EXAMPLES   FOK   PRACTICE.  375 


Ex.15.  Given  a^z  =  105, 

xyv  —  135,      to   find  the  values   of  #,  y,  z, 
xzv  —  189,  and  v. 


Ans.  #=3,  y=§i  z  =  7,  v  =  9. 

Ex.16.  Given  x2  +  -9+y2=84:J] 

y2  '  I    to   find   the   values   of  as 

x2  and  y. 

*+-  +  y=14. 

U  J 

Ans.  cc  =  4,  y  =  2  or  8. 

Ex.  17.  Given  Vy—  Va—x         =  Vy—x,   1  to  find  the  val- 
2Vy—x  +  2Va—x  =  oVa—x,j  uesofxandy. 


Ans.  x  —   a     = 


Ex.18.  Given  x3 


=  ay.}        „    ,    .  _ 

^      v  to  find  the  values  of  x  and  y. 
=  bx,  j 


Ex.19.  GivenA/5-v/x  +  5\/?/4-  Vx+  Vy=   10,  /  to   find  the 
V^5+  Vy~5  =275,  )  values  of  x 

and  y. 

Remark.  Put  z*=-\/x  +  -\fy.  Then,  from  Eq.  1,  z=y~5;  that  is,  -\/lc  +  Vy  =  5. 
Next  put  *\/x=^+v,  and  ~Vy—^— v.  Substituting  these  values  in  Eq.  2,  we 
find  u2=i,  or  v=±J. 

-13 

.  a?  =  U  or  4,  or  — 


=4  or  9,  or 


Several  of  the  following  examples  have  imaginary  or  incom 
mensurable  roots  which  are  not  here  given. 

Ex.20.  Given  (x2  +  y2)xy  =  13090,  )  to  find  the  values  of  as 
x+y        —       18,  )  and  y. 

Ans.  x~l  or  11,  y  =  ll  or  7. 

Ex.  21.  Given  5(x2+?/2)+4x?/  =356,  )  to  find  the  values  of 

'=   62,  )  x  and  y. 

Ans.  x  =  4  or  6,  y  —  6  or  4. 


376  ALGEBRA. 

Ex.22.  Given  (x2^y*)xy=SOO, )    to  find  the  values  of  on 
x*  +  y*      r=337,  )  and  y. 

Ans.  x—  ±4,  y—  ±3. 

Ex.  23.  Given  (x*+y2)(x3+y3)=4:55,  )  to  find  the  values  of 
x+y  —     5,  )  x  and  y. 

Ans.  oc  —  3  or  2,  y  =  2  or  3. 

Ex.24.  Given  _,..,, 

to  nna  the  values   of   x 

and  y. 

'-y 

Ans.  cc=12,  y  =  6. 

Ex.25.  Given  (tf—xy+y^^+y*)         =  91,  )  to  find  the 
(x*  —  xy+y2)(x2+xy-{-y2)  =  133,  )  values  of  x 
and?/.  J-ws.  cc=  ±3  or  ±2,  ?/=  ±2  or  ±3. 

Ex.  26.  Given  (x+y)xy      —  30,  )  to  find  the  values  of  x 

!=468,  )  and  y. 

Ans.  x—2  or  3,  y  =  3  or  2. 
/x y       ^2    1 

Ex.27.  Given  cc—v+V -= »1    to  nn(^  tne  values 

v  -  '  -    x-|-?/  >  „ 

'  ^   1  of  a?  and  y. 

:41,        j 

:=±5,  y=±4. 


Ex.28.  Given  (x+y)3+x+y=30,  }  to  find  the  values  of  x 
x—y=  1,  )  and  ?/. 

Remark.  Multiply  Eq.  1  by  ar+y,  and  we  have 


Add  to  each  member  9(a;+#)2  +  25,  and  the  square  root  of  each  member  of  the 
equation  may  be  extracted. 

Ans.  x=2,  y=l. 

Ex.29.  Given  (x  +y)(xy  +1)=   ISxy    )  to  find  the  vat 
(x2+7/2)(xY+l)  =  208xy  )    ues  of  a?  and  y. 

Remark.  Divide  Eq.  1  by  xy,  and  Eq.  2  by  a;2//2,  and  we  have 

x  +y  +-  +- 
x     y 


Put  x+-=z,  and  «+- 
x 


-  +-  =  18. 


Then  z+v  =  18,  and  z2+ua=212. 

Whence  2=14  or  4,  and  t?=4  or  14  ;  and  hence  x  and  y  are  easily  found 

Ans.  x-2±  -v/3,  y=7 


EXAMPLES   FOE   PKACTICE.  377 

PROBLEMS  INVOLVING-  EQUATIONS  OF  THE  SECOND  DEGREE 
WITH   SEVERAL   UNKNOWN   QUANTITIES. 

Prob.  1.  If  I  increase  the  numerator  of  a  certain  fraction  by 
2,  and  diminish  the  denominator  by  2, 1  obtain  the  reciprocal 
of  the  first  fraction ;  also,  if  I  diminish  the  numerator  by  2,  and 
increase  the  denominator  by  2,  the  resulting  fraction,  increased 
by  1^,  is  equal  to  the  reciprocal  of  the  first  fraction.  What 
is  the  fraction  ?  Ans.  y. 

Prob.  2.  It  is  required  to  divide  the  number  102  into  three 
parts,  such  that  the  product  of  the  first  and  third  shall  be  equal 
to  102  times  the  second  part,  and  the  third  part  shall  be  1-J 
times  the  first. 

Ans.  The  first  part  is  34,  the  second  17,  and  the  third  51. 

Prob.  3.  A  certain  number  consists  of  two  digits.  If  I  in 
vert  the  digits,  and  multiply  this  new  number  by  the  first,  I  ob 
tain  for  a  product  5092 ;  but  if  I  divide  the  first  digit  by  the 
second,  I  obtain  1  for  a  quotient  with  1  for  a  remainder.  What 
is  the  number?  Ans.  76. 

Prob.  4.  The  fore  wheel  of  a  carriage  makes  165  more  rev 
olutions  than  the  hind  wheel  in  going  5775  feet ;  but  if  the 
circumference  of  each  wheel  be  increased  2-J-  feet,  the  fore 
wheel  will  make  only  112  revolutions  more  than  the  hind 
wheel  in  the  same  space.  Required  the  circumference  of  each 
wheel. 

Ans.  The  fore  wheel  10  feet,  the  hind  wheel  14  feet. 

Prob.  5.  A  piece  of  cloth,  by  being  wet  in  water,  shrinks  one 
eighth  in  its  length  and  one  sixteenth  in  its  breadth.  If  the 
perimeter  of  the  piece  is  diminished  4J  feet,  and  the  surface  5f 
square  feet,  by  wetting,  what  were  the  length  and  breadth  of 
the  piece? 

Ans.  16  feet  long  and  2  feet  wide. 

Prob.  6.  A  certain  number  of  laborers  in  8  hours  transport  a 
pile  of  stones  from  one  place  to  another.  If  there  were  8  more 
laborers,  and  if  each  carried  each  time  5  pounds  less,  the  pile 
would  be  removed  in  7  hours ;  but  if  there  were  8  less  labor 
ers,  and  if  each  carried  each  time  11  pounds  more,  it  would  re- 


'378  ALGEBKA. 

quire  9  hours  to  remove  the  pile.     How  many  laborers  were 
there  employed,  and  how  many  pounds  did  each  carry  ? 

Ans.  28  laborers,  and  each  carried  45  pounds  ;  or  36 

laborers,  and  each  carried  77  pounds.     -f£. 
Prob.  7.  A  certain  capital  yields  yearly  $123-J-  interest  ;   a 
second  capital,  $700  larger,  and  loaned  at  J  per  cent,  less,  yields 
yearly  $11-1  more  interest  than  the  first.     How  large  was  the 
first  capital,  and  at  what  per  cent,  was  it  loaned  ? 

Ans.  The  capital  was  $3800,  loaned  at  3J  per  cent. 
Prob.  8.  A  person  bought  a  number  of  $20  railway  shares 
when  they  were  at  a  certain  rate  per  cent,  discount  for  $1500  ; 
and  afterward,  when  they  were  at  the  same  rate  per  cent,  pre 
mium,  sold  them  all  but  60  for  $1000.  How  many  did  he  buy, 
and  what  did  he  give  for  each  of  them  ? 

Ans.  100  shares  at  $15  each. 

Prob.  9.  A  rectangular  lot  is  119  feet  long  and  19  feet  broad. 
How  much  must  be  added  to  the  breadth,  and  how  much  taken 
from  the  length,  in  order  that  the  perimeter  may  be  increased 
by  24  feet,  and  the  contents  of  the  lot  remain  the  same  ? 

Ans.  The  length  must  be  diminished  102  feet,  and  the 

breadth  increased  114  feet. 

Prob.  10.  There  are  two  numbers  such  that  their  sum  and 
product  together  amount  to  47  ;  also,  the  sum  of  their  squares 
exceeds  the  sum  of  the  numbers  themselves  by  62.  What  are 
the  numbers? 

Ans.  5  and  7. 

Prob.  11.  The  sum  of  two  numbers  is  a,  and  the  sum  of  their 
reciprocals  is  b.  Kequired  the  numbers. 


Prob.  12.  A  and  B  engage  to  reap  a  field  for  $24  ;  and  as  A 
alone  could  reap  it  in  nine  days,  they  promise  to  complete  it  in 
five  days.  They  found,  however,  that  they  were  obliged  to 
call  in  C  to  assist  them  for  the  last  two  days,  in  consequence 
of  which  B  received  one  dollar  less  than  he  otherwise  would 
have  done.  In  what  time  could  B  or  C  alone  reap  the  field? 

Ans.  B  in  15  and  C  in  18  days. 


EXAMPLES  FOR  PEACTICE.  379 

Prob.  13.  The  sum  of  the  cubes  of  two  numbers  is  35,  and 
the  sum  of  their  ninth  powers  is  20,195.  Kequired  the  numbers. 

Ans.  2  and  3. 

Prob.  14.  There  are  two  numbers  whose  product  is  300 ;  and 
the  difference  of  their  cubes  is  thirty-seven  times  the  cube  ol 
their  difference.  What  are  the  numbers  ? 

Ans.  20  and  15. 

Prob.  15.  A  merchant  had  $26,000,  which  he  divided  into  two 
parts,  and  placed  them  at  interest  in  such  a  manner  that  the  in 
comes  from  them  were  equal.  If  he  had  put  out  the  first  por 
tion  at  the  same  rate  as  the  second,  he  would  have  drawn  for 
this  part  $720  interest ;  and  if  he  had  placed  the  second  out  at 
the  same  rate  as  the  first,  he  would  have  drawn  for  it  $980  in 
terest.  What  were  the  two  rates  of  interest? 

Ans.  6  per  cent,  for  the  larger  sum,  and  7  for  the  smaller. 

Prob.  16.  A  miner  bought  two  cubical  masses  of  ore  for  $820. 
Each  of  them  cost  as  many  dollars  per  cubic  foot  as  there  were 
feet  in  a  side  of  the  other;  and  the  base  of  the  greater  contain 
ed  a  square  yard  more  than  the  base  of  the  less.  What  was 
the  price  of  each?  Ans.  500  and  320  dollars. 

Prob.  17.  A  gentleman  bought  a  rectangular  lot  of  land  at 
the  rate  of  ten  dollars  for  every  foot  in  the  perimeter.  If  the 
same  quantity  had  been  in  a  square  form,  and  he  had  bought  it 
at  the  same  rate,  it  would  have  cost  him  $330  less ;  but  if  he 
had  bought  a  square  piece  of  the  same  perimeter,  he  would  have 
had  12J  rods  more.  What  were  the  dimensions  of  the  lot  ? 

AJIS.  9  by  16  rods. 

Prob.  18.  A  and  B  put  out  at  interest  sums  amounting  to 
$2400.  A's  rate  of  interest  was  one  per  cent,  more  than  B's; 
his  yearly  interest  was  five  sixths  of  B's ;  and  at  the  end  of  ten 
years  his  principal  and  simple  interest  amounted  to  five  sev 
enths  of  B's.  What  sum  was  put  at  interest  by  each,  and  at 
what  rate? 

^ir      Ans.  A  $960  at  5  per  cent,  B  $1440  at  4  per  cent. 

Prob.  19.  A  person  bought  a  quantity  of  cloth  of  two  sorts 
for  $63.  For  every  yard  of  the  best  piece  he  gave  as  many 
dollars  as  he  had  yards  in  all ;  and  for  every  yard  of  the  poor- 


380  ALGEBRA. 

er,  as  many  dollars  as  there  were  yards  of  the  better  piece  more 
than  of  the  poorer.  Also,  the  whole  cost  of  the  best  piece  was 
six  times  that  of  the  poorer.  How  many  yards  had  he  of  each  ? 

Ans.  6  yards  of  the  better  and  3  of  the  poorer. 
Prob.  20.  A  commences  a  piece  of  work  alone,  and  labors 
for  two  thirds  of  the  time  that  B  would  have  required  to  per; 
form  the  entire  work.  B  then  completes  the  job.  Had  both 
labored  together,  it  would  have  been  completed  two  days  soon 
er,  and  A  would  have  performed  only  half  what  he  left  for  B. 
Eequired  the  time  in  which  they  would  have  performed  the 
work  separately. 

Remark.  Suppose  A  would  have  performed  the  work  in  x  days,  and  B  in  y  days. 

A  labors  -J-  days,  and  performs  —  part  of  the  work. 
o  ox 

O0I         O~» O*» 

B  performs  1  —-^-= — - — -  part  of  the  work. 
ox         ox 

3x—2y 

— - — -xy= time  B  labored. 

• — — +—  =whole  time  consumed. 

ox  o 

-+-~part  both  did  in  one  day— -. 

x     y  xy 

— — =the  days  of  work  if  both  labored  together. 

xy    ,9_3xy—2yz2y 
x+y  3x         ~3 ' 

Also,  ix-^-= iof°~ 


*"*+i     2          3x    ' 

Ans.  A  in  6  days  and  B  in  3  days. 


EXAMPLES   FOR  PRACTICE.  381 


PROGRESSIONS. 

Ex.  1.  What  is  the  sum  of  the  natural  series  of  numbers 
1,  2,  3,  etc.,  up  to  1000? 

Ans.  500,500. 

Ex.  2.  What  is  the  sum  of  an  arithmetical  progression  whose 
first  term  is  6,  the  last  term  2833,  and  the  number  of  terms  38? 

Ans.  53,941. 

Ex.  3.  What  is  the  first  term  and  the  sum  of  the  terms  of  an 
arithmetical  progression,  when  the  last  term  is  24,  the  common 
difference  -f-,  and  the  number  of  terms  22  ? 

Ans.  First  term  9,  and  sum  of  terms  363. 
Ex.  4.  Kequired  the  number  and  the  sum  of  the  terms  of  an 
arithmetical  progression,  when  the  first  term  is  —  f ,  the  com 
mon  difference  — |,  and  the  last  term  —2 If. 

Ans.  Number  of  terms  25,  and  sum  of  terms  —  281 J. 
Ex.  5.  The  first  term  of  an  arithmetical  progression  is  5,  the 
last  term  23,  and  the  sum  of  the  terms  392.     What  is  the  com 
mon  difference  and  the  number  of  terms  ? 

Ans.  Common  difference  •§,  and  number  of  terms  28. 
Ex.  6.  Between  7  and  13  it  is  required  to  interpolate  8  terms 
which  shall  form  an  arithmetical  progression. 

Ans.  7f,  8t,  9,  9|,  IQfr,  11,  llf,  12f 

Ex.  7.  In  an  arithmetical  progression,  the  sum  of  the  19th, 
the  43d,  and  the  57th  terms  is  827;  the  sum  of  the  27th,  the 
58th,  the  69th,  and  the  73d  terms  is  1581.  What  is  the  first 
term  and  the  common  difference  ? 

-4.W.9.  The  first  term  is  5,  and  the  common  difference  7 
Ex.  8.  In  boring  an  artesian  well  500  feet  deep,  $3.24  is  paid 
for  the  first  foot,  and  5  cents  more  for  each  subsequent  foot. 
How  much  was  paid  for  the  last  foot,  and  how  much  for  the 
whole  well  ? 

Ans.  For  the  last  foot  $28.19,  and  for  the  entire  well 

$7857i 

Ex.  9.  According  to  natural  philosophy,  a  body  falling  in  a 
vacuum  describes  in  the  first  second  of  its  fall  16^  feet,  and 


382  ALGEBRA. 

in  each  succeeding  second  32^  feet  more  than  in  the  second 
immediately  preceding.  If  a  body  has  fallen  20  seconds,  how 
many  feet  will  it  fall  in  the  last  second,  and  how  many  in  the 
whole  time? 

Ans.  627J  feet  in  the  last  second,  and  6433^  feet  in 
the  whole  time. 

^  Ex.  10.  Divide  unity  into  four  parts  in  arithmetical  progres 
sion,  of  which  the  sum  of  the  cubes  shall  be  VT-. 


Ex.  11.  A  ship,  with  a  crew  of  175  men,  set  sail  with  a  sup 
ply  of  water  sufficient  to  last  to  the  end  of  the  voyage ;  but  in 
30  days  the  scurvy  made  its  appearance,  and  carried  off  three 
men-  every  day;  and  at  the  same  time  a  storm  arose,  which 
protracted  the  voyage  three  weeks.  They  were,  however, 
just  enabled  to  arrive  in  port  without  any  diminution  in  each 
man's  daily  allowance  of  water.  Eequired  the  time  of  the 
passage,  and  the  number  of  men  alive  when  the  vessel  reached 
the  harbor. 

Ans.  The  voyage  lasted  79  days,  and  the  number  of 
men  alive  was  28. 

Ifamark.  Put  x = days  the  voyage  was  expected  to  last. 

a;-f-21=days  the  voyage  lasted. 

a;  +  21-30=a;-9=the  days  after  30. 
On  the  31st  day  the  number  of  men  was  172,  etc. 

Last  term       =  172-3(o;-10). 

Sum  of  series =(344- 3(o:  - 10))  x^?. 

Then  (344  -3* +  30)^5  =  175  (x  -30). 

Whence  x =58.     Also,  x +21  =  79  days  the  voyage  lasted. 

Ex.  12.  The  number  of  deaths  in  a  besieged  garrison  amount 
ed  to  6  daily ;  and,  allowing  for  this  diminution,  their  stock  of 
provisions  was  sufficient  to  last  8  days.  But  on  the  evening 
of  the  sixth  day  100  men  were  killed  in  a  sally,  and  afterward 
the  mortality  increased  to  10  daily.  Supposing  the  stock  of 
provisions  unconsumed  at  the  end  of  the  sixth  day  to  support  6 
men  for  61  days,  it  is  required  to  find  how  long  it  would  sup- 


EXAMPLES  FOR  PRACTICE.  383 

port  the  garrison,  and  the  number  of  men  alive  when  the  pro 
visions  were  exhausted. 

Ans.  The  provisions  last  6  days,  and  26  men  survive. 

Remark.  Put  a:— number  of  men  at  first. 

x— 42=number  expected  at  end  of  8  days. 

9r_  49 

— -x8  =  8x  — 168  = number  of  days'  provisions. 

2 

___X6  =  6:E  —  90=days'  provisions  exhausted  at  end  of  6th  day. 
2 

2X—  78  =366  =the  remainder. 
Whence  #=222. 

222  — 136  =  86— number  of  men  after  the  sally. 
Put  y=number  of  days  the  provisions  lasted  afterward. 


Ex.  13.  The  first  term  of  a  geometrical  progression  is  1,  the 
ratio  2,  arid  the  number  of  terms  13.  What  is  the  last  term 
and  the  sum  of  the  terms  ? 

Ans.  The  last  term  is  4096,  and  the  sum  of  the  terms 

8191. 

Ex.  14.  The  first  term  of  a  geometrical  progression  is  7,  the 
ratio  3,  and  the  number  of  terms  11.  What  is  the  last  term 
and  the  sum  of  the  terms  ? 

Ans.  The  last  term  is  413,343,  and  the  sum  of  the 

terms  620,011. 

Ex.  15.  The  sum  of  the  terms  of  a  geometrical  progression 
is  411,771,  the  ratio  7,  and  the  number  of  terms  7.  Eequired 
the  first  and  last  terms. 

Ans.  First  term  3,  and  last  term  352,947. 
Ex.  16.  Between  1  and  £  it  is  required   to  interpolate  11 
terms  forming  a  continued  geometrical  progression.     What  are 
the  terms? 

Ans.  0.9439,  0.8909,  0.8409,  0.7937,  0.7492,  0.7071, 

0.6674,  0.6300,  0.5946:  0.5612,  0.5297. 

Ex.  17.  What  will  $1200  amount  to  in  36  years  at  4  per 
cent,  compound  interest?  Ans.  $4924.70. 

Ex.  18.  A  farmer  sowed  a  peck  of  wheat,  and  used  the  whole 
crop  for  seed  the  following  year;  the  produce  of  the  second 
year  he  uaed  for  seed  the  third  year,  and  so  on.  If  in  the  10th 


884  ALGEBRA. 

year  the  crop  was  1,048,576  pecks,  by  how  many  times  must  the 
seed  have  increased  each  harvest^  supposing  the  increase  to 
have  been  always  the  same  ?  Ans.  Four  times. 

Ex.  19.  There  are  three  numbers  in  geometrical  progression^ 
the  difference  of  whose  differences  is  six,  and  their  sum  is  forty- 
two.  Kequired  the  numbers.  Ans.  6, 12,  and  24. 

Ex.  20.  There  are  three  numbers  in  geometrical  progression, 
the  greatest  of  which  exceeds  the  least  by  24 ;  and  the  differ 
ence  of  the  squares  of  the  greatest  and  the  least  is  to  the  sum 
of  the  squares  of  all  the  three  numbers  as  5  :  7.  What  are  the 
numbers?  Ans.  8,  16,  and  32. 

Ex.  21.  There  are  three  numbers  in  geometrical  progression 
whose  continued  product  is  216,  and  the  sum  of  their  cubes  is 
1971.  Kequired  the  numbers.  Ans.  8,  6,  and  12. 

Ex.  22.  There  are  four  numbers  in  geometrical  progression 
whose  sum  is  850 ;  and  the  difference  between  the  extremes  is 
to  the  difference  of  the  means  as  87 : 12.  What  are  the  num 
bers?  Ans.  54,  72,  96,  128. 

Ex.  23.  There  are  four  numbers,  the  first  three  of  which  are 
in  geometrical  progression,  and  the  last  three  in  arithmetical 
progression  ;  the  sum  of  the  first  and  last  is  14,  and  that  of  the 
second  and  third  12  ;  find  the  numbers. 

25  15  9  3 
Ans.  2,  4,  8,  12 ;  or  —  — ,  -,  -. 

Ex.  24.  Three  numbers  whose  sum  is  15  are  in  arithmetical 
progression ;  if  1,  4,  and  19  be  added  to  them  respectively,  they 
are  in  geometrical  progression.  Determine  the  numbers. 

Ans.  2,  5,  8. 
qfc. 


THE  END. 


cr     /Vwvx^    ,,        &\ 


< 


. 


LOAN  PERIOD  1 

2                               3 

4 

5                                6 

ALL  BOOKS  MAY  BE  RECALLED  AFTER  7  DAYS 

DUE  AS  STAMPED  BELOW 

iN  1  bKUBR/ 

^RY  LOAN 

JAN  3  ' 

'1989 

UNIV.  OF  CA 

LIR.  eg&tf, 

UNIVERSITY  OF  CALIFORNIA,  BERKELEY 
FORM  NO.  DD07  15m,  2/84          BERKELEY,  CA  94720 


10 


924206 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


iff 


